The definitions of a convergent sequence, a Cauchy sequence and a complete space, together with the three facts about real sequences used everywhere later: the Cauchy criterion, monotone convergence, and the existence of limsup and liminf.
NoteOriginal
What it means. Convergence asks for a limit to be named; the Cauchy condition asks only that the terms crowd together, which is something you can check without knowing the answer. Completeness is the statement that in this space the second test never lies, and it is what makes R usable for analysis. Monotone convergence is the cheapest way to produce a limit (bound a monotone quantity and you are done), and limsup and liminf are the fallback when a sequence need not converge: they always exist, so one can argue with them and only afterwards show that they agree.
Definition(Convergent sequence)AdaptedAdapted from Kreyszig (1978), §1.4, Def. 1.4-1 and Lemma 1.4-2, pp. 25–26; Reed & Simon (1980), §I.2, p. 5
Let (X,d) be a metric space and let (xn)n∈N be a sequence of points of X. The sequence converges to x∈X when d(xn,x)→0; written out, each ε>0 has an index N∈N, which may depend on ε, such that d(xn,x)<ε at every n>N. Such an x is unique if it exists, and is called the limit, written xn→x or limn→∞xn=x. A sequence with no limit in X is divergent. Throughout, R and C carry the metric d(x,y)=∣x−y∣, for which this is the notion of convergence used in calculus.
Definition(Cauchy sequence, complete space)AdaptedAdapted from Kreyszig (1978), §1.4, Def. 1.4-3, p. 28; Reed & Simon (1980), §I.2, pp. 5–8
Call (xn) in (X,d) a Cauchy sequence when its terms eventually crowd together: each ε>0 admits an index N beyond which d(xm,xn)<ε holds for every pair m,n>N. Only the terms enter this condition; no candidate limit is mentioned.
Every convergent sequence is Cauchy, since d(xn,x)<ε/2 for n>N gives d(xm,xn)≤d(xm,x)+d(x,xn)<ε for m,n>N. The space (X,d) is complete when the converse holds throughout it, so that each of its Cauchy sequences has a limit lying in X.
Definition(Supremum, monotone sequences, limsup and liminf)AdaptedAdapted from Kreyszig (1978), App. A1.6, pp. 619–620; Folland (1999), §0.5, pp. 10–11
A non-empty set E⊆R that is bounded above has a least upper bound supE∈R, and one bounded below has a greatest lower bound infE; this least-upper-bound property is the order-theoretic form of the completeness of R. In the extended reals R=R∪{−∞,+∞} every subset has a supremum and an infimum, with supE=+∞ when E is unbounded above and infE=−∞ when it is unbounded below.
A real sequence (xn) is bounded if {xn:n∈N} is bounded above and below, non-decreasing if xn≤xn+1 for every n, and non-increasing if xn+1≤xn for every n; it is monotone if it is one of these. Its limit superior and limit inferior are
both computed in R. They exist for every real sequence, since k↦supn≥kxn is non-increasing and k↦infn≥kxn is non-decreasing.
Definition(Cauchy criterion in R and C)AdaptedAdapted from Kreyszig (1978), §1.4, Thm 1.4-4, p. 28 and App. A1.7, Cauchy convergence theorem, p. 621
A real sequence has a limit precisely when it is Cauchy, and the same holds for a complex sequence: ∣xm−xn∣ can be forced below any prescribed ε>0 for all large m and n exactly when the sequence converges. Equivalently, R and C with d(x,y)=∣x−y∣ are complete metric spaces.
Idea of the proof(Cauchy criterion)Original
The condition does two jobs. It confines the sequence: once all late terms are within 1 of each other they are within 1 of one fixed term, so the sequence is bounded. Boundedness, through the least-upper-bound property, produces a point approached by infinitely many terms — the Bolzano–Weierstrass theorem, which we prove here by halving an interval repeatedly, since the map carries it in no unit that precedes this one. The Cauchy condition then does its second job: a term far out is close to the terms of that subsequence, which are close to the point, so the whole sequence converges. The complex case reduces to the real one by taking real and imaginary parts.
Proof(Cauchy criterion)AdaptedAdapted from Kreyszig (1978), App. A1.7, p. 621, with App. A1.6, pp. 619–620
Necessity is the implication recorded with the definition above, which used only the triangle inequality. We prove sufficiency, first for a real Cauchy sequence (xn).
Step 1: a Cauchy sequence is bounded. Take ε=1 in the Cauchy condition: some N has ∣xm−xn∣<1 for all m,n>N. Holding n=N+1 fixed, ∣xm∣≤∣xN+1∣+1 for every m>N. The remaining indices 1,…,N are finite in number, so
C=max{∣x1∣,…,∣xN∣,∣xN+1∣+1}
is a real number and ∣xn∣≤C for every n.
Step 2 (lemma: Bolzano–Weierstrass on the line).A real sequence with ∣xn∣≤C for all n has a subsequence converging to a point of [−C,C]. Set [a0,b0]=[−C,C], an interval containing xn for infinitely many n — indeed for all of them. Given an interval [ak,bk] with that property, cut it at its midpoint: the indices n with xn∈[ak,bk] split between the two closed halves, and two finite sets cannot have an infinite union, so at least one half contains xn for infinitely many n. Name it [ak+1,bk+1]. By construction ak≤ak+1≤bk+1≤bk and bk−ak=21−kC.
The set {ak:k≥0} is non-empty and bounded above by b0, so the least-upper-bound property gives α=supkak∈R. Each bk is itself an upper bound of that set, because aj≤ak≤bk for j≤k and aj≤bj≤bk for j≥k; hence α≤bk, and ak≤α by definition, so α∈[ak,bk] for every k, and in particular α∈[−C,C].
Choose indices n1<n2<⋯ with xnk∈[ak,bk], which is possible because each of these intervals holds xn for infinitely many n, so an index larger than those already chosen is always available. Both xnk and α lie in [ak,bk], whence
∣xnk−α∣≤bk−ak=21−kC⟶0,
so xnk→α. This proves the lemma.
Step 3: from the subsequence to the sequence. Let α be the limit just produced and let ε>0. The Cauchy condition supplies an N with ∣xm−xn∣<ε/2 for all m,n>N, and xnk→α supplies one index nk>N with ∣xnk−α∣<ε/2. For every m>N the triangle inequality gives
∣xm−α∣≤∣xm−xnk∣+∣xnk−α∣<ε.
So xm→α, and the real case is settled.
Step 4: the complex case. Let (zn) be Cauchy in C and write zn=pn+iqn with pn,qn real. Since ∣pm−pn∣≤∣zm−zn∣ and ∣qm−qn∣≤∣zm−zn∣, both real sequences are Cauchy, so by Steps 1–3 they converge, say pn→p and qn→q. Then ∣zn−(p+iq)∣≤∣pn−p∣+∣qn−q∣→0, so zn→p+iq.
Convergent sequences are Cauchy in any metric space, so both implications hold in R and in C, and each is therefore complete. □
Definition(Monotone convergence)Original
Let (xn) be a non-decreasing sequence of real numbers and S={xn:n∈N}.
If S is bounded above, then (xn) converges and limn→∞xn=supS.
If S is not bounded above, then for every M∈R there is an N with xn>M for all n>N; one writes xn→+∞ and the sequence is divergent.
For a non-increasing sequence the mirror statements hold, with infS in place of supS and −∞ in place of +∞. In every case limn→∞xn=supnxn for non-decreasing and infnxn for non-increasing sequences, as a limit in R.
Idea of the proof(Monotone convergence)Original
A non-decreasing sequence has nowhere to go but up towards its supremum. The least-upper-bound property names that supremum L, and "least" means that nothing below L bounds the sequence, so some term already exceeds L−ε; monotonicity then holds every later term in the thin strip between L−ε and L. If there is no upper bound, the same remark applied to an arbitrary level M shows that the terms pass M and never return.
Proof(Monotone convergence)Original
Let (xn) be non-decreasing and put S={xn:n∈N}, a non-empty subset of R.
Bounded above. The least-upper-bound property gives L=supS∈R. Fix ε>0. Because L is the least upper bound, L−ε is not an upper bound, so xN>L−ε for at least one index N. For n>N, monotonicity gives xn≥xN>L−ε, while xn≤L because L bounds S. Hence ∣xn−L∣<ε for all n>N. As ε>0 was arbitrary, xn→L=supS.
Not bounded above. Let M∈R. Then M fails to be an upper bound of S, so xN>M for some N, and monotonicity gives xn≥xN>M for all n>N. That is the assertion xn→+∞. Such a sequence has no real limit, because a convergent sequence is bounded: all but finitely many of its terms lie within 1 of the limit, and the finitely many others have a largest absolute value.
Non-increasing sequences. If (xn) is non-increasing then (−xn) is non-decreasing, and sup{−xn:n}=−inf{xn:n}; since −xn→ℓ is equivalent to xn→−ℓ, parts 1 and 2 applied to (−xn) give the mirror statements, with infS in place of supS and −∞ in place of +∞. □
Definition(Convergence through limsup and liminf)AdaptedAdapted from Folland (1999), §0.5, p. 11
Every real sequence satisfies liminfnxn≤limsupnxn in R. The sequence converges to a real limit if and only if
n→∞liminfxn=n→∞limsupxnand this common value is finite,
and then it is the limit.
Idea of the proof(Convergence through limsup and liminf)Original
Only the monotonicity of the two auxiliary sequences ak=supn≥kxn and bk=infn≥kxn is needed. The inequality liminf≤limsup comes from comparing bj with ak at the later of the two indices, where one is an infimum and the other a supremum of the same tail. If the sequence converges, every late term sits in an interval of width 2ε about the limit, and so therefore do ak and bk. Conversely, if the two numbers agree at a finite value x, then one tail supremum lies below x+ε and one tail infimum above x−ε, and together they squeeze all late terms to within ε of x.
Proof(Convergence through limsup and liminf)Original
Write ak=supn≥kxn and bk=infn≥kxn, computed in R, where every set has a supremum and an infimum; thus limsupnxn=infkak and liminfnxn=supkbk by definition.
The inequality. Fix indices j and k and let l=max(j,k). Then bj≤bl, because the infimum over the smaller tail {xn:n≥l} can only be larger; bl≤al, because that tail is non-empty; and al≤ak, because the supremum over a smaller tail can only be smaller. So bj≤ak for every pair j,k. Taking the supremum over j gives liminfnxn≤ak for each k, and then the infimum over k gives liminfnxn≤limsupnxn.
Convergence forces equality at a finite value. Suppose xn→x∈R and let ε>0. Some N has x−ε≤xn≤x+ε for all n>N. For k>N the whole tail {xn:n≥k} lies in [x−ε,x+ε], so ak≤x+ε and bk≥x−ε. Hence limsupnxn≤ak≤x+ε and liminfnxn≥bk≥x−ε, which with the inequality already proved gives
x−ε≤n→∞liminfxn≤n→∞limsupxn≤x+ε.
Since ε>0 was arbitrary, both are equal to x, and in particular both are finite.
Equality at a finite value forces convergence. Suppose liminfnxn=limsupnxn=x with x∈R, and let ε>0. From infkak=x<x+ε there is a K1 with aK1<x+ε, so xn<x+ε for all n≥K1. From supkbk=x>x−ε there is a K2 with bK2>x−ε, so xn>x−ε for all n≥K2. With N=max(K1,K2) we get ∣xn−x∣<ε for all n≥N. Hence xn→x, the limit being the common value. □
Completeness is a property of the space, not of the sequence. Take X=Q with d(x,y)=∣x−y∣ and a sequence of rationals converging in R to an irrational number: it is Cauchy in Q and has no limit in Q. So "Cauchy implies convergent" does not follow from the metric axioms; for R it is a theorem, resting on the least-upper-bound property through the Bolzano–Weierstrass theorem (see Compactness).
Monotone convergence needs the bound.xn=n is non-decreasing with no real limit. It also needs the least-upper-bound property, and not merely monotonicity and boundedness: the non-decreasing rational sequence of n-digit decimal truncations of 2 is bounded above by 2 and has no limit in Q.
limsup and liminf need no hypotheses, but may be infinite. Equality of the two alone is not convergence: for xn=n both equal +∞. Finiteness is what the criterion above adds.
Boundedness is necessary but not sufficient for convergence, as xn=(−1)n shows, with liminf=−1 and limsup=1.
Example(A convergent sequence that no monotonicity argument reaches)Original
Define x1=1 and xn+1=1/(1+xn). The first terms are
1,21,32,53,85,138,2113,…
— consecutive ratios of Fibonacci numbers — and they fall and rise alternately, so monotone convergence has nothing to say. The Cauchy criterion does.
Every term lies in [21,1]. True for x1=1. If xn∈[21,1] then 1+xn∈[23,2], so xn+1=1/(1+xn)∈[21,32]⊂[21,1]. Induction does the rest.
Consecutive differences shrink by a factor 94. For n≥2,
the last step being the sum of a geometric series with ratio 94. The right-hand side does not depend on m and tends to 0, so given ε>0 any n with 109(4/9)n−1<ε serves as the required index. By the Cauchy criterion the sequence converges; call the limit L, and L∈[21,1] by step 1.
Identify L. The map t↦1/(1+t) is continuous at L, so xn+1=1/(1+xn)→1/(1+L); but xn+1→L as well, and limits are unique, so L(1+L)=1. Of the two roots of L2+L−1=0 only one is positive, so
L=25−1=0.6180339887…
Check.x10=55/89=0.6179775280…, which differs from L by 5.6×10−5, comfortably inside the bound 109(4/9)9=6.1×10−4 that step 3 gives for ∣xm−x10∣.
Example(A bounded sequence with liminf=limsup)Original
Every term satisfies ∣xn∣≤2, so the sequence is bounded.
Fix k and let e(k) be the smallest even integer ≥k. Among the terms with n≥k the positive ones are 1+1/n for even n, which decrease as n grows, and the negative ones are below 0; so the largest is the one at n=e(k) and
ak=n≥ksupxn=1+e(k)1.
Since e(k)→∞ and every ak>1, we get limsupnxn=infkak=1. The same argument with the smallest odd integer o(k)≥k gives bk=infn≥kxn=−(1+1/o(k)) and liminfnxn=supkbk=−1.
For example a5=1+61=67 (the supremum is attained at n=6) and b5=−56 (attained at n=5). The two limits are finite but unequal, so by the criterion above the sequence does not converge, which the subsequences confirm: x2m→1 and x2m+1→−1.
Application(Solving Kepler's equation by iteration)Original
A body on an elliptical orbit of eccentricity e∈[0,1) has its position fixed by the eccentric anomaly E, which solves Kepler's equation
E−esinE=M,
where the mean anomaly M grows linearly in time. There is no elementary formula for E, and the oldest numerical method is the fixed-point iteration
E0=M,Ek+1=M+esinEk.
The iteration map has derivative ecosE, of either sign, so depending on M the iterates may climb steadily towards the root or oscillate about it; monotonicity is not available in general, and the limit is not known in advance either, so there is nothing to compare the terms against. What is available is the increment between successive terms, and that is exactly what the Cauchy criterion asks for. By the mean value theorem applied to sin,
∣Ek+1−Ek∣=e∣sinEk−sinEk−1∣≤e∣Ek−Ek−1∣,
because ∣cos∣≤1; hence ∣Ek+1−Ek∣≤ek∣E1−E0∣ and, summing the geometric series as in the example above, ∣Em−En∣≤en∣E1−E0∣/(1−e) for m>n. The eccentricity being below 1 is precisely what makes this tend to 0: the iteration is Cauchy, the completeness of R supplies a limit E, and continuity of sin turns the recursion into E=M+esinE in the limit.
With e=0.2 and M=1 (radians), ∣E1−E0∣=esin1=0.1682942, so the bound predicts ∣E−E8∣≤0.28×0.1682942/0.8=5.4×10−7. The iteration gives E=1.1853242039, with E−esinE−M below 10−14. For comets, with e close to 1, the factor en decays too slowly and the same completeness argument, while still valid, is useless in practice; Newton's method is used instead.
Partial sums, the vanishing-terms and Cauchy criteria, and the comparison, ratio, root, integral and alternating-series tests, each with the hypotheses it needs and the cases in which it decides nothing.
What it means. Every test here is a comparison with a series whose behaviour is already known: the geometric series for the ratio and root tests, an integral for the integral test, and an explicit comparand for the comparison test. So the practical question is not "which test is strongest" but "which known series does mine resemble": geometric decay calls for the ratio or root test, power-law decay for comparison with a p-series, and anything on the boundary between them for the integral test.
Let (un)n≥1 be real or complex numbers, and write
sN=n=1∑Nun,N∈N,
for the partial sums. The symbol ∑n=1∞un denotes the series, and every question about it is a question about the sequence (sN). The series converges to S when sN→S, and S is then its sum; otherwise the series diverges. For a convergent series the tail or remainder after N terms is RN=S−sN=∑n>Nun. Deleting, inserting or changing finitely many terms changes S but not whether the series converges, so every test below may be applied from any index onwards.
Let ∑n≥1un be a series of real or complex terms.
Cauchy criterion. Convergence holds exactly when every block of consecutive terms far enough out is small: each ε>0 admits an index N with
n=m+1∑m+pun<εwhenever m≥N and p≥1.
Vanishing terms. If the series converges then un→0. The converse is false, so un→0 is necessary and not sufficient; if un→0 the series diverges.
Idea of the proof(Two criteria that need no extra hypotheses)Original
Both are the corresponding facts about the sequence (sN) in disguise. The Cauchy condition on partial sums is a condition on their differences, and a difference of two partial sums is exactly a block of consecutive terms. Taking the shortest block, of one term, turns that condition into the vanishing of the terms.
Proof(Two criteria that need no extra hypotheses)AdaptedAdapted from Ahlfors (1979), §2.2, p. 35
Cauchy criterion. By definition the series converges exactly when the sequence (sN) converges, and by the Cauchy criterion for numerical sequences that happens exactly when each ε>0 has an index K with ∣sj−si∣<ε for all i,j>K. Every such pair with i<j can be written as i=m, j=m+p with m≥N:=K+1 and p≥1, and conversely; and
sm+p−sm=n=m+1∑m+pun.
Pairs with i=j contribute nothing and pairs with i>j give the same modulus. So the displayed condition of part 1 is the Cauchy condition on (sN), which completes the equivalence.
Vanishing terms. Take p=1 in part 1: convergence gives an N with ∣um+1∣<ε for every m≥N, which says un→0. (Equivalently, un=sn−sn−1→S−S=0.) The converse fails: the p-series with p=1 has un=1/n→0 and diverges, as the integral test proved below shows. Contrapositively, un→0 forces divergence. □
If un≥0 for all n, the partial sums are non-decreasing, so by monotone convergence the series converges if and only if its partial sums are bounded above, and then ∑n≥1un=supNsN. This is the fact behind every comparison below, and it is why a series of non-negative terms either converges or has sN→+∞ and can never oscillate.
Idea of the proof(Non-negative terms)Original
Adding a non-negative number cannot decrease a partial sum, so the partial sums are a monotone sequence, and for monotone sequences boundedness and convergence are the same thing.
Proof(Non-negative terms)Original
Since sN+1−sN=uN+1≥0 for every N, the sequence (sN) is non-decreasing. By monotone convergence it converges if its terms are bounded above, with limit supNsN, and otherwise sN→+∞. Convergence of the series is convergence of (sN), so the two conditions match, and the sum is supNsN.
Only these two behaviours are available: a non-decreasing sequence that is bounded above converges and one that is not tends to +∞, so the partial sums of a series of non-negative terms cannot oscillate. □
In each part, N denotes an index from which the stated condition holds; nothing is assumed about the finitely many earlier terms.
Comparison test. Suppose 0≤un≤vn for all n≥N. If ∑vn converges then ∑un converges. Suppose instead 0≤vn≤un for all n≥N. If ∑vn diverges then ∑un diverges.
Ratio test (d'Alembert). Suppose un>0 for all n≥N. If there is a constant r<1, independent of n, with un+1/un≤r for all n≥N, then ∑un converges. If un+1/un≥1 for all n≥N, then ∑un diverges. In limit form, if ρ=limn→∞un+1/un exists, then ∑un converges for ρ<1, diverges for ρ>1, and the test decides nothing for ρ=1.
Root test (Cauchy). Suppose un≥0 for all n≥N. If there is a constant r<1, independent of n, with un1/n≤r for all n≥N, then ∑un converges. If un1/n≥1 for all n≥N, then ∑un diverges. In limit form, if ρ=limn→∞un1/n exists, then ∑un converges for ρ<1, diverges for ρ>1, and the test decides nothing for ρ=1.
Integral test (Cauchy-Maclaurin). Let f be continuous, non-negative and non-increasing on [1,∞) and let un=f(n). Then ∑n≥1un converges if and only if ∫1∞f(x)dx is finite, and in that case, for every N≥0,
∫N+1∞f(x)dx≤n>N∑un≤∫N+1∞f(x)dx+uN+1.
Alternating-series test (Leibniz). Let un≥0 with un→0, and suppose un+1≤un for all n≥N. Then ∑n≥1(−1)n+1un converges, its sum S lies between any two consecutive partial sums sm and sm+1 with m≥N, and consequently
∣S−sm∣≤um+1,
with S−sm having the sign of the first omitted term (−1)mum+1.
Reference series. The geometric series ∑n≥0rn converges exactly for ∣r∣<1, and the p-series ∑n≥1n−p converges exactly for p>1; these are the standard comparands.
Idea of the proof(The five tests)Original
Tests 1 to 4 all end at the same place: a series of non-negative terms whose partial sums have been bounded above, which converges by the statement just proved. What differs is the bound used — the comparand itself, a geometric series in the ratio and root tests, an integral in the integral test. The divergence halves are cheaper still: each either exhibits terms that do not tend to 0, or pushes the partial sums past every level. The Leibniz test is of a different kind: there the partial sums of even and of odd index form two monotone sequences that close in on each other, and the gap between them is a single term.
Assume 0≤un≤vn for n≥N and that ∑vn converges. All terms vn with n≥N are non-negative, being at least the corresponding un, so the tail V=∑n≥Nvn is a finite number that dominates each of its own partial sums. Hence for every M≥N
n=N∑Mun≤n=N∑Mvn≤V,
term by term in the first inequality. The partial sums of ∑n≥Nun are therefore bounded above, and its terms are non-negative, so it converges by the statement above. Restoring the finitely many terms u1,…,uN−1 changes the sum but not the convergence.
For the second half, suppose 0≤vn≤un for n≥N and that ∑vn diverges. If ∑un converged, the first half — applied with the roles of the two series exchanged — would make ∑vn converge, a contradiction. So ∑un diverges. □
Both tests compare with a geometric series, which is why both are silent when the decay is slower than geometric.
Root test, convergence. Raising un1/n≤r to the n-th power, which preserves the inequality between non-negative numbers, gives un≤rn for n≥N. With 0≤r<1 the series ∑nrn converges, so the comparison test applies.
Root test, divergence. If un1/n≥1 for n≥N then un≥1 for those n, so un→0 and the vanishing-terms criterion gives divergence.
Ratio test, convergence. The hypothesis un+1≤run for n≥N, iterated k times, gives uN+k≤uNrk; the hypothesis un>0 is what makes the ratios meaningful and the induction legitimate. Since ∑kuNrk=uN/(1−r) converges for 0≤r<1, the comparison test again gives convergence.
Ratio test, divergence. If un+1/un≥1 for n≥N then un≥uN>0 for all n≥N, so the terms cannot tend to 0.
Limit forms. Suppose ρ=limnun+1/un exists and ρ<1. Put r=(1+ρ)/2, so that ρ<r<1; by the definition of the limit, applied with ε=r−ρ, there is an N with un+1/un<r for all n≥N, and the convergent half applies with this constant r. If ρ>1, the same definition with ε=ρ−1 gives an N with un+1/un>1 for n≥N, and the divergent half applies. The root test is identical with un1/n in place of un+1/un. The case ρ=1 is left undecided, and the remark below shows that nothing better is possible. □
Proof(Integral test and its tail bound)AdaptedAdapted from Arfken, Weber & Harris (2013), §1.1, pp. 5-6, Eqs. (1.7) and (1.9)
One interval at a time. Fix an integer n≥1. For x∈[n,n+1] the hypothesis that f is non-increasing gives f(n+1)≤f(x)≤f(n), and integrating these constants and f over an interval of length 1 gives
un+1≤∫nn+1f(x)dx≤un,
the integral existing because f is continuous.
Summing. Let N≥0 and M>N be integers. Adding the last display over n=N+1,…,M and using additivity of the integral,
n=N+2∑M+1un≤∫N+1M+1f(x)dx≤n=N+1∑Mun.
The equivalence. Because f≥0, the numbers IM=∫1Mf form a non-decreasing sequence, so ∫1∞f=limMIM exists in [0,+∞] by monotone convergence. Take N=0 above. If ∫1∞f is finite, then ∑n=2M+1un≤∫1M+1f≤∫1∞f for every M: the partial sums are bounded, the terms are non-negative since f≥0, and the series converges. If ∫1∞f=+∞, then ∑n=1Mun≥∫1M+1f→+∞ and the series diverges. The two cases are exclusive and exhaustive, which is the stated equivalence.
The tail bound. Assume convergence and let M→∞ in the summed display, at fixed N≥0; all three quantities increase to their limits, so
n>N+1∑un≤∫N+1∞f(x)dx≤n>N∑un.
The right-hand inequality is the lower bound asserted in the statement. Adding uN+1 to the left-hand one and using ∑n>Nun=uN+1+∑n>N+1un gives the upper bound. □
Write sm=∑n=1m(−1)n+1un and let m≥N, so that the monotonicity hypothesis is available from index m+1 on.
One step.sm+1−sm=(−1)m+2um+1=(−1)mum+1, which is ≥0 for even m and ≤0 for odd m.
Two steps.sm+2−sm=(−1)m(um+1−um+2), and um+1≥um+2 by monotonicity, so this is ≥0 for even m and ≤0 for odd m. Hence, from index N onwards, the partial sums of even index form a non-decreasing sequence and those of odd index a non-increasing one.
The two families interleave. Let m≥N be even and m′≥N be odd, and pick an even l≥max(m,m′). Then sm≤sl (even family, non-decreasing), sl≤sl+1 (one step from an even index), and sl+1≤sm′ (odd family, non-increasing, and l+1>m′). So every even-index partial sum is at most every odd-index one.
Both families converge, to the same number. The even family is non-decreasing and bounded above (by any odd-index sum), and the odd family is non-increasing and bounded below, so by monotone convergence both converge, say to Se≤So. Along even m, sm+1−sm=um+1→0 by the hypothesis un→0, and the left-hand side tends to So−Se; hence Se=So=:S. Given ε>0, each family is eventually within ε of S, and the larger of the two thresholds serves all indices, so sm→S and the series converges with sum S.
Where S sits. For even m≥N the even family increases to S and the odd family decreases to S, so sm≤S≤sm+1; for odd m≥N the same reading gives sm+1≤S≤sm. Either way S lies between the two consecutive partial sums, whence
∣S−sm∣≤∣sm+1−sm∣=um+1,
and S−sm has the sign of sm+1−sm=(−1)mum+1, which is the first omitted term. □
Geometric series. For r=1, multiplying sN=∑n=0Nrn by 1−r cancels all but two terms and gives sN=(1−rN+1)/(1−r). If ∣r∣<1, write ∣r∣=1/(1+h) with h>0; then (1+h)N+1≥1+(N+1)h→∞, so ∣r∣N+1→0 and sN→1/(1−r). If ∣r∣≥1 then ∣rn∣≥1 for every n, the terms do not tend to 0, and the series diverges (this covers r=1, where the partial sums are N+1).
The p-series. For p≤0 we have n−p≥1, so the terms do not vanish and the series diverges. For p>0 the function f(x)=x−p is continuous, positive and non-increasing on [1,∞) with f(n)=n−p, so the integral test applies to it:
∫1Mx−pdx=1−pM1−p−1(p=1),∫1Mx−1dx=lnM.
As M→∞ the first tends to 1/(p−1) when p>1 and to +∞ when 0<p<1, and the second to +∞. So the integral is finite exactly for p>1, and so is the series. □
Method(Deciding whether a series converges)Original
Check the terms. If un→0, stop: the series diverges. This catches ∑n/(n+1) and ∑(−1)n at once.
Fix the sign. If the terms have one sign from some index on, work with ∣un∣ and go to step 3. If the signs alternate strictly, go to step 6. If the signs change irregularly, test ∑∣un∣ by steps 3-5: convergence there settles the question (see Absolute and conditional convergence), and divergence there does not.
Compare. Find the size of un for large n and compare with a geometric or p-series: if un≤Cn−p with p>1 for large n, the series converges; if un≥c/n with c>0 for large n, it diverges. This is decisive whenever un behaves like a power of n times a bounded factor.
Ratio or root. If un contains factorials or products, form un+1/un; if it contains an n-th power, form un1/n. A limit ρ<1 gives convergence and ρ>1 divergence. These tests work by hidden comparison with a geometric series, so they are decisive exactly when the terms decay geometrically.
Integrate. If un=f(n) for an f that is continuous, non-negative and non-increasing on [1,∞), use the integral test. It settles the borderline cases that step 4 cannot, such as ∑1/(nlnn), and its two-sided bound also gives a numerical estimate of the tail.
Alternating terms. For ∑(−1)n+1un with un≥0, check both Leibniz hypotheses: un→0 and un eventually non-increasing. The error bound ∣S−sm∣≤um+1 comes free.
If a test fails, nothing follows.ρ=1 in step 4, or a failed comparison, says only that this test is silent; return to step 3 with a sharper comparand, or to step 5.
Remark(Where the hypotheses are used, and sharper forms)Original
Non-negativity in tests 1-4. All four rest on bounded monotone partial sums, which needs the terms to have one sign. Applied to a series with mixed signs, they test ∑∣un∣ instead, which is a strictly stronger property.
The ratio test needs a constant r<1, not just un+1/un<1. For the harmonic series un+1/un=n/(n+1)<1 for every n, yet no fixed r<1 dominates these ratios and the series diverges; the limit form gives ρ=1, which decides nothing.
ρ=1 is genuinely undecided.∑n−2 converges and ∑n−1 diverges, and both have ρ=1 in the ratio and root tests.
Monotonicity in the integral test. It is what makes un+1≤∫nn+1f≤un hold, which is where the two-sided bound comes from. Without it, f(n)=un constrains f only at the integers and the integral can be made to converge or diverge at will.
Monotonicity in the Leibniz test cannot be dropped, even though un→0: take un=1/n for odd n and un=1/n2 for even n. Then un≥0 and un→0, but ∑n≥1(−1)n+1un=∑odd n1/n−∑even n1/n2 has partial sums tending to +∞, because the first sum diverges and the second converges.
Sharper divergence halves. The divergence conditions in tests 2 and 3 can be weakened, at the cost of departing from the form in which Arfken states them (pp. 4-5): if un1/n≥1 for infinitely many n, then un≥1 infinitely often, so un→0 and the series already diverges. Likewise, the tests remain valid with limsup in place of lim on the convergent side and liminf on the divergent side, because the condition "un+1/un≤r<1 for all large n" is exactly limsupnun+1/un<1.
Comparison from an index onwards. Arfken states the comparison test with 0≤un≤an for all n (p. 3); Riley, Hobson and Bence state it for n greater than some fixed N (p. 125). The second is the more general form and is the one stated above; the two agree because finitely many terms cannot affect convergence.
Example(The ratio test, and then the exact sum)Original
Decide whether n≥1∑3nn3 converges, and if so sum it.
Step 1: convergence. With un=n33−n>0,
unun+1=3n3(n+1)3=31(1+n1)3,
which decreases as n grows. At n=3 its value is 31(4/3)3=64/81, so
unun+1≤r=8164<1for all n≥3,
a constant bound as the ratio test requires, and the series converges. (The limit form gives ρ=1/3, the same conclusion.) Note how the test is used. The ratio exceeds 1 at n=1 and n=2 — its values there are 8/3 and 9/8 — and that costs nothing, because the hypothesis is only ever imposed from some index onwards. What the test does need is a single r<1 above the ratio for all those n, which here is any number in [64/81,1).
Step 2: the sum. Put Sk=∑n≥1nk3−n for k=0,1,2,3; each converges by the computation above with nk in place of n3. Multiplying a convergent series by 3 and shifting the index by one (which only relabels the partial sums),
the last step splitting off the m=0 term. Subtracting Sk, and using that convergent series may be subtracted termwise,
2Sk=1+m≥1∑3m(m+1)k−mk.
Now take k=0,1,2,3 in turn. For k=0 every bracket is 0, so 2S0=1 and S0=21, which is the geometric series ∑m≥13−m as it should be. For k=1, (m+1)−m=1, so 2S1=1+S0=23 and S1=43. For k=2, (m+1)2−m2=2m+1, so 2S2=1+2S1+S0=1+23+21=3 and S2=23. For k=3, (m+1)3−m3=3m2+3m+1, so
Check. The first twenty terms already give 4.1249986, so the answer 4.125 is right to the digits shown.
Example(A tail estimated with the integral test)Original
The series ∑n≥1n−2 is a p-series with p=2>1, so it converges. Suppose we have summed ten terms,
s10=n=1∑10n21=1.5497677…,
and want to know what is missing. Take f(x)=x−2, which is continuous, positive and non-increasing on [1,∞) with f(n)=n−2, and apply the two-sided bound of the integral test with N=10. Since ∫11∞x−2dx=1/11 and u11=1/121,
The bracket has width u11=1/121, so its midpoint 1/11+1/242=0.0950413 estimates the tail with an error of at most 1/242=4.1×10−3. Adding it to s10 gives 1.6448090 for the sum of the series.
Check. The exact tail is R10=0.0951663…, inside the bracket as promised, and the midpoint rule missed it by only 1.2×10−4. (The classical value of the whole series is π2/6=1.6449341…; it is not needed for the estimate and is quoted only to check it.) Ten terms plus one integral thus do the work of several thousand terms of the bare series, which is the practical value of the integral test's bound.
Application(Convergence of a partition function)Original
In statistical mechanics the equilibrium properties of a system follow from its partition function, a sum over states whose convergence is a physical question, not a formality. For a quantum harmonic oscillator of angular frequency ω the energies are En=ℏω(n+21) with n=0,1,2,…, and at inverse temperature β=1/kBT
Z(β)=n≥0∑e−βEn=e−x/2n≥0∑(e−x)n,x=βℏω.
The ratio of consecutive terms is the constant e−x, so the ratio test settles the matter at once: the sum converges for every x>0 and diverges for x≤0. Being geometric, it can then be evaluated:
Z=1−e−xe−x/2=ex/2−e−x/21=2sinh(βℏω/2)1,
from which the mean energy follows as ⟨E⟩=−∂βlnZ=21ℏωcoth(βℏω/2). At βℏω=0.7, for instance, Z=1.3998163.
Two features of the convergence are physics. First, the condition x>0 says that the partition function of a system with infinitely many ever-higher levels exists only at positive temperature; the divergence at β=0 is the statement that infinite temperature would populate infinitely many states equally. Second, the geometric decay means the sum converges at a rate set by e−βℏω, and truncating it after N terms leaves
n>N∑e−βEn=1−e−βℏωe−βℏω(N+3/2),
a bound of the kind one needs for the anharmonic and rotational sums that have no closed form: there the comparison is made against this geometric series, and the number of levels to keep is read off from the required accuracy.
A series whose absolute values converge is convergent and its sum is unchanged by any rearrangement; a convergent series of real terms that is not absolutely convergent can be rearranged to any sum whatever, and the Cauchy product of two absolutely convergent series converges to the product of their sums.
What it means. Absolute convergence is the licence to treat an infinite sum like a finite one: reorder it, split it, multiply it by another such sum. Conditional convergence gives you a number and nothing else, and for a real series Riemann's theorem says exactly how little: the number itself is an artefact of the order in which you added the terms. In practice, before reordering a sum, multiplying two sums or exchanging a sum with another limit, establish absolute convergence; that is what the tests of the previous unit are for.
Let ∑n≥0an be a series of real or complex terms. It is absolutely convergent if ∑n≥0∣an∣ converges. It is conditionally convergent if it converges but ∑n≥0∣an∣ does not.
Because ∑∣an∣ has non-negative terms, the comparison, ratio, root and integral tests of Convergence tests for series are tests for absolute convergence.
A rearrangement of ∑n≥0an is the series ∑n≥0aσ(n) for a bijection σ:N0→N0, where N0={0,1,2,…}.
The Cauchy product of ∑n≥0an and ∑n≥0bn is the series ∑n≥0cn with
cn=k=0∑nakbn−k,
that is, the series obtained by collecting the products akbl with k+l=n. It is the series one gets by multiplying two power series term by term and gathering equal powers.
Theorem(Absolute convergence implies convergence)AdaptedAdapted from Ahlfors (1979), §2.2, p. 35; Hassani (2013), §10.6.1, p. 319
If ∑n≥0∣an∣ converges, with an real or complex, then ∑n≥0an converges. The converse fails, so absolute convergence is strictly stronger than convergence.
Idea of the proof(Absolute convergence implies convergence)Original
Neither series' sum is known, so the only usable test is the Cauchy criterion, which asks about blocks of consecutive terms rather than about a limit. A block of the given series is dominated in modulus by the corresponding block of the series of absolute values, so smallness of the second kind of block is inherited by the first.
Proof(Absolute convergence implies convergence)AdaptedAdapted from Ahlfors (1979), §2.2, p. 35
the modulus signs outside the sum being unnecessary because the terms are non-negative. The triangle inequality for a finite sum then gives
n=m+1∑m+pan≤n=m+1∑m+p∣an∣<ε
for the same m and p. That is the Cauchy criterion for ∑n≥0an, which therefore converges. In Ahlfors's phrasing, the partial sums of ∑an are a contraction of those of ∑∣an∣: they move no further apart, so they inherit the Cauchy property.
The converse fails: ∑n≥1(−1)n−1/n converges by the Leibniz test while ∑n≥11/n diverges, so absolute convergence is strictly stronger. □
Theorem(Bound on the sum)Original
If ∑n≥0an is absolutely convergent, with an real or complex, then
n=0∑∞an≤n=0∑∞∣an∣,
with equality if and only if there is a constant λ∈C with ∣λ∣=1 such that λan=∣an∣ for every n; that is, if and only if all the non-zero terms have the same argument. For real terms this says that the terms are all ≥0 or all ≤0.
The bound is the triangle inequality for the partial sums carried to the limit on both sides. It is not printed on either of the pages cited for the previous statement, so it is stated here as the handbook's own.
Idea of the proof(Bound on the sum)Original
The inequality is the triangle inequality for each partial sum, kept through the limit because a non-strict inequality survives it. For the equality case, rotate the whole series by the unit number λ that makes its sum real and positive. Then ∑Re(λan) and ∑∣an∣ have the same value, while the terms of the first never exceed those of the second; a series of non-negative terms with sum zero has every term zero, so the two agree term by term, and a complex number whose real part equals its modulus is non-negative.
Proof(Bound on the sum)Original
Write sN=∑n=0Nan, S=∑n=0∞an — which exists by the statement just proved — and T=∑n=0∞∣an∣.
The inequality. The triangle inequality for finite sums gives ∣sN∣≤∑n=0N∣an∣≤T, the second step because the partial sums of a series of non-negative terms never exceed its sum. The modulus is continuous, so ∣sN∣→∣S∣, and a non-strict inequality is preserved in the limit: ∣S∣≤T.
Equality implies a common argument. Suppose ∣S∣=T.
The degenerate case. If T=0, then the partial sums of ∑∣an∣ are non-decreasing with limit 0, so each is both ≥0 and ≤0, hence 0; so every ∣an∣=0 and λ=1 satisfies λan=∣an∣ for every n.
The main case. If T>0 then S=0, and we may set λ=S/∣S∣, so that ∣λ∣=1 and λS=∣S∣=T. Multiplying a convergent series by a constant and taking real parts termwise are both limits of the corresponding operations on partial sums, so
n=0∑∞Re(λan)=Re(λS)=T.
Each term satisfies Re(λan)≤∣λan∣=∣an∣, so the series ∑n(∣an∣−Re(λan)) has non-negative terms, and its sum is T−T=0. By the degenerate case's argument every term vanishes:
Re(λan)=∣an∣=∣λan∣for every n.
A complex number z with Rez=∣z∣ satisfies (Imz)2=∣z∣2−(Rez)2=0, so it is real and equal to its own modulus, hence ≥0. Applied to z=λan this gives λan=∣λan∣=∣an∣ for every n, which is the asserted condition.
A common argument implies equality. Conversely, if ∣λ∣=1 and λan=∣an∣ for every n, then summing gives λS=T, so ∣S∣=∣λS∣=T.
Reading the condition.λan=∣an∣ with ∣λ∣=1 says an=λ∣an∣, so every non-zero term has argument −argλ: the non-zero terms share an argument. If the terms are real and some an=0, then λ=∣an∣/an=±1, and the condition says that all terms are ≥0 (for λ=1) or all ≤0 (for λ=−1). □
Let ∑n≥0an be absolutely convergent with sum S. Then for every bijection σ of N0 the rearranged series ∑n≥0aσ(n) is again absolutely convergent and its sum is S. So the order of summation is immaterial once absolute convergence is known.
Idea of the proof(Rearrangement of an absolutely convergent series)Original
Absolute convergence means that all but finitely many terms together contribute less than ε in total. Once a partial sum of the rearranged series is long enough to contain all of those finitely many terms, whatever else it contains is drawn from the negligible remainder. So both the original sum and every long enough rearranged partial sum sit within ε of the same finite sum, and hence within 2ε of each other.
Proof(Rearrangement of an absolutely convergent series)Original
Let S=∑n≥0an, let T=∑n≥0∣an∣, let σ be a bijection of N0, and let ε>0.
A negligible tail. The partial sums of ∑∣an∣ increase to T, so there is an N with ∑n>N∣an∣=T−∑n=0N∣an∣<ε.
An index after which nothing is missing. Each of the finitely many integers 0,1,…,N has exactly one σ-preimage, because σ is a bijection. Put M=max{σ−1(0),…,σ−1(N)}. For every m≥M the list σ(0),…,σ(m) contains 0,…,N, and its entries are distinct.
Comparison of partial sums. For m≥M the difference
j=0∑maσ(j)−n=0∑Nan
is a sum of finitely many distinct terms an, all with n>N; by the triangle inequality its modulus is at most ∑n>N∣an∣<ε. The original series satisfies the same kind of bound: by the statement proved above, applied to the series ∑n>Nan,
S−n=0∑Nan=n>N∑an≤n>N∑∣an∣<ε.
Adding the two estimates, ∑j≤maσ(j)−S<2ε for every m≥M. Since ε>0 was arbitrary, ∑j≥0aσ(j) converges with sum S.
The rearrangement is again absolutely convergent. Every partial sum ∑j≤m∣aσ(j)∣ consists of finitely many distinct terms of the non-negative series ∑∣an∣, so it is at most T. Non-negative terms and bounded partial sums give convergence, and the argument just given — applied to ∑∣an∣ in place of ∑an — shows that the sum is again T. □
Let ∑n≥0an be a series of real terms that converges but is not absolutely convergent. Then:
for every L∈R there is a bijection σ of N0 with ∑n≥0aσ(n)=L;
there are bijections for which the partial sums tend to +∞, and others for which they tend to −∞;
there are bijections for which the partial sums oscillate and have no limit in R.
So for a conditionally convergent series the sum is a property of the ordering and not of the set of terms alone.
Idea of the proof(Riemann's rearrangement theorem)Original
Split the terms into the non-negative ones and the negative ones. Neither list can have a finite total: if both did, the series would converge absolutely, and if one did, the other would too. So each list is an unlimited supply of material of one sign, and the terms are individually small because the series converges. Drawing on the two lists in turn, taking from each only as much as is needed to overshoot a target and then to undershoot it, steers the partial sums wherever one pleases: towards a chosen number, to infinity, or back and forth between two levels.
Proof(Riemann's rearrangement theorem)AdaptedAdapted from Arfken, Weber & Harris (2013), §1.1, Example 1.1.9, pp. 14-15
Throughout, ∑an converges, so an→0, and ∑∣an∣ diverges.
Lemma: the two signs separately diverge. Put an+=max(an,0) and an−=max(−an,0), both non-negative, with an=an+−an− and ∣an∣=an++an−. Since an+=an+an− and ∑an converges, the series ∑an+ and ∑an− converge or diverge together. They cannot both converge, for then ∑∣an∣=∑(an++an−) would converge. So both diverge, and because their terms are non-negative their partial sums tend to +∞.
The two lists. List the terms an≥0, in their original order, as p1,p2,…, and the moduli of the terms an<0, in their original order, as q1,q2,…. Each list is infinite: if, say, only finitely many terms were negative, ∑an− would be a finite sum. Deleting the zero entries changes nothing, so ∑kpk=∑nan+=+∞ and ∑kqk=∑nan−=+∞, while pk→0 and qk→0 because an→0. Every term of the original series occupies exactly one position in exactly one list, so any prescription that consumes both lists in order, each entry once and all of them eventually, defines a bijection σ of N0, that is, a rearrangement.
Part 1: an arbitrary real limit L. Build the rearrangement in stages, each of which takes at least one entry from its list and then continues until a target is crossed. Stage 1 ascends: add p1,p2,… in order until the running total has reached L or more, which happens after finitely many entries because ∑pk=+∞. Stage 2 descends: add −q1,−q2,… in order until the total has fallen below L, which happens because ∑qk=+∞. Stage 3 resumes with the next unused p, and so on. The stages alternate forever and each uses at least one entry, so both lists are consumed in order and the prescription is a rearrangement.
Let vk be the total at the end of stage k and tk the last entry it used. Each stage adds entries of one sign, so the partial sums occurring inside stage k lie between vk−1 and vk.
From stage 2 onwards each stage starts on the far side of L from the direction it moves: an ascending stage starts below L, because the preceding descending stage ended there, and a descending stage starts at or above L. Such a stage therefore ends just after its first crossing of L: for an ascending one, L≤vk≤L+∣tk∣, since one entry earlier the total was still below L; for a descending one, L−∣tk∣≤vk<L. So ∣vk−L∣≤∣tk∣ for every k≥2.
As k grows both lists advance without bound, so ∣tk∣→0 because pk→0 and qk→0. Every partial sum occurring in stage k, for k≥3, therefore differs from L by at most max(∣tk−1∣,∣tk∣), which tends to 0; the finitely many partial sums of the first two stages are irrelevant to a limit. Hence the rearranged series converges to L.
Part 2: divergence to ±∞. For +∞, let stage k take unused entries p in order — at least one — until the total exceeds k, and then exactly one unused entry −q. Each stage terminates and uses entries of both lists, so this is again a rearrangement. Write qjk for the single negative entry of stage k. The total at the end of stage k exceeds k−qjk, and inside stage k+1 the total only grows until that stage's final entry; so every partial sum from the end of stage k onwards is at least k−supj≥jkqj. Since qj→0, this bound tends to +∞ with k, and so do the partial sums. Exchanging the roles of the two lists gives −∞.
Part 3: oscillation. Alternate two targets: take unused p's — at least one — until the total exceeds 1, then unused q's until it falls below 0, and repeat forever. As before this is a rearrangement. The partial sums exceed 1 infinitely often and fall below 0 infinitely often, so they have no limit. They are also bounded: an ascending stage starts at or below 0 and stops within one entry of 1, and a descending stage starts at most at 1+P and stops within one entry of 0, so every partial sum lies in [−Q,1+P], where P=supkpk and Q=supkqk are finite because both lists tend to 0. So the partial sums do not tend to +∞ or −∞ either, and have no limit in R. □
Let ∑n≥0an and ∑n≥0bn be absolutely convergent, with sums A and B. Then their Cauchy product ∑n≥0cn, cn=∑k=0nakbn−k, is absolutely convergent and
n=0∑∞cn=AB.
The same holds for the terms akbl summed in any other order, since the product series converges absolutely.
Idea of the proof(Cauchy product)Original
Think of all the products akbl as a single infinite list. Absolute convergence of the factors bounds the sum of their moduli by (∑∣ak∣)(∑∣bl∣), so that list is an absolutely convergent series and, by the rearrangement theorem, has a sum independent of the order chosen. Two orders are then read off: along expanding squares, where the partial sums are ANBN; and along diagonals, where they are the partial sums of the Cauchy product. Both must give the same number.
Proof(Cauchy product)Original
Write α=∑k≥0∣ak∣ and β=∑l≥0∣bl∣, both finite, and AN=∑k=0Nak, BN=∑l=0Nbl.
Step 1: the products form one absolutely convergent series. Fix any bijection j↦(kj,lj) from N0 onto N0×N0 and consider the series ∑j≥0akjblj. Given J, let K be the largest of the numbers k0,l0,…,kJ,lJ. The pairs (kj,lj) with j≤J are distinct and all lie in the square {0,…,K}2, so
the first inequality because every product on the left appears exactly once among the non-negative products obtained by expanding the brackets. Non-negative terms with bounded partial sums converge, so ∑jakjblj is absolutely convergent; call its sum T. Two enumerations of N0×N0 differ by a bijection of N0, so by the rearrangement theorem T does not depend on which one was chosen.
Step 2: enumerate by squares to get AB. List first the pair (0,0), then the pairs with max(k,l)=1, then those with max(k,l)=2, and so on, in any fixed order inside each block. The partial sum at the end of the block max(k,l)=N runs over all pairs in {0,…,N}2, so it equals ANBN. These are partial sums of a convergent series with sum T, so ANBN→T; but AN→A and BN→B, so ANBN→AB. Hence T=AB.
Step 3: enumerate by diagonals to get the Cauchy product. List instead the pairs with k+l=0, then those with k+l=1, and so on. The partial sum at the end of the block k+l=N runs over all pairs with k+l≤N, so it equals ∑n=0Ncn. Again these are partial sums of a series with sum T, so ∑n=0Ncn→T=AB.
Step 4: the product series converges absolutely. By the triangle inequality ∣cn∣≤∑k+l=n∣ak∣∣bl∣, and summing over n≤N gives ∑n=0N∣cn∣≤αβ by the bound of Step 1. Non-negative terms with bounded partial sums converge.
So ∑ncn is absolutely convergent with sum AB, and Step 1 is the final sentence of the statement: the products akbl summed in any order give the same T=AB. □
Both rearrangement results need their hypothesis, and the two are complementary. The alternating harmonic series ∑n≥1(−1)n−1/n converges by the Leibniz test and diverges in absolute value; Arfken exhibits a reordering of it (taking positive terms until the partial sum reaches 3/2, then negative terms until it falls below, and repeating) whose partial sums converge to 3/2 instead of to ln2 (p. 15).
The Cauchy product needs more than convergence of the factors. Take an=(−1)n/n+1, so that ∑n≥0an converges conditionally. Its Cauchy product with itself has
cn=(−1)nk=0∑n(k+1)(n−k+1)1,
and the sum on the right is a Riemann sum for ∫01t(1−t)dt=π, so ∣cn∣→π. The terms do not tend to 0, so the product series diverges, as Arfken's Example 1.1.10 (p. 15) observes.
Only one factor need be absolutely convergent. If ∑an converges absolutely to A and ∑bn converges to B, the Cauchy product still converges to AB; this is Mertens' theorem, and Ahlfors sets it as Exercise 6 (p. 37), asking for a proof arranged so that the second factor need only converge. The conclusion is then weaker: the product series converges to AB but need not converge absolutely.
The restriction to real terms in Riemann's theorem is ours, not the books'. Both cited sources state the theorem for any conditionally convergent series and attach no condition to the terms. Arfken draws it as a moral from the rearranged alternating harmonic series: a suitable reordering can send such a series to any value one likes, or make it diverge (p. 15). Riley, Hobson and Bence list the three possible behaviours — an arbitrary limit, divergence, finite or infinite oscillation — two sentences after saying that the terms of a series may be complex (p. 124). Read as printed over C, that is false: if an=i(−1)n/(n+1), the series converges and is not absolutely convergent, yet every term is purely imaginary, so every convergent rearrangement has a purely imaginary sum and no real L=0 is attainable. Part 1 above is therefore restricted to real terms, which is where both books' examples live; the two printed statements are logged in verification/discrepancies.md.
What replaces part 1 for complex terms. A theorem of Levy and Steinitz: the sums attainable by convergent rearrangements of a convergent complex series form a translate of a linear subspace of C. A single point is exactly the absolutely convergent case, since in finite dimensions unconditional convergence forces absolute convergence; so for a series that converges but not absolutely the set of attainable sums is a line or the whole of C. Over R the only proper subspace is {0}, and the theorem collapses to Riemann's.
Terminology. "Conditionally convergent" here means convergent but not absolutely convergent, following Arfken (p. 13) and Riley, Hobson and Bence (p. 124). Some texts use the phrase for any convergent series whose convergence depends on the arrangement of its terms, which for real series is the same class.
Example(Rearranging a conditionally convergent series halves it)Original
The alternating harmonic series ∑n≥1(−1)n−1/n converges by the Leibniz test, and ∑1/n diverges, so it is conditionally convergent and the rearrangement theorem does not apply to it. Call its sum Λ. Rearrange it by taking one positive term and then two negative ones, over and over:
1−21−41+31−61−81+51−101−121+⋯
Every odd reciprocal appears once, in its original order, and every even reciprocal appears once with a minus sign, in its original order, so this is a genuine rearrangement.
Step 1: the k-th group. It is
gk=2k−11−4k−21−4k1.
Since 2k−11−4k−21=2(2k−1)1=4k−21, this simplifies to
gk=4k−21−4k1=21(2k−11−2k1).
Step 2: sum the groups. The bracket is a consecutive pair of terms of the original series, so adding g1,…,gK gives exactly half the 2K-th partial sum of the original series:
k=1∑Kgk=21n=1∑2Kn(−1)n−1⟶2Λ.
Step 3: fill in the partial sums between group boundaries. A partial sum of the rearranged series differs from the nearest group boundary by at most two of its terms, each of modulus at most 1/(4k−2)→0, so the whole rearranged series converges to Λ/2.
Check. With Λ=ln2=0.6931472 (quoted only as a reference value), the rearranged sum should be 0.3465736. Adding the first 200000 groups gives 0.3465730, and the remaining tail is about 1/(8K)=6×10−7, which accounts for the difference.
The same terms, reordered, sum to half as much. Riemann's theorem says that Λ/2 is nothing special: some rearrangement of these terms converges to each real number.
Example(A Cauchy product evaluated)Original
Take an=2−n and bn=3−n for n≥0. Both series have positive terms and converge, hence converge absolutely, with
A=n≥0∑2−n=2,B=n≥0∑3−n=23.
The product terms. Summing a finite geometric series with ratio 3/2,
The first values are c0=1, c1=5/6, c2=19/36, c3=65/216, which agree with the definition computed directly.
The sum. Both geometric series again,
n≥0∑cn=3n≥0∑2−n−2n≥0∑3−n=3⋅2−2⋅23=3=AB,
as the theorem predicts, and the convergence is absolute because cn>0 (indeed 3/2n>2/3n for every n≥0).
Example(When the triangle-inequality bound is attained)Original
Let an=eiπ/42−n for n≥0. Then ∑n≥0∣an∣=∑2−n=2 and ∑n≥0an=2eiπ/4, whose modulus is also 2: the bound is an equality. The equality condition is satisfied with λ=e−iπ/4, since λan=2−n=∣an∣ for every n — all the terms share the argument π/4.
Reverse the sign of one term, say replace a1 by −a1. The series of moduli is unchanged, so the right-hand side is still 2, while the sum loses 2a1=eiπ/4 and becomes eiπ/4, of modulus 1. Now no unit λ can satisfy λan=∣an∣ for both n=0 and n=1, because those two terms have arguments differing by π; and indeed 1<2 strictly.
Application(Lattice sums in a crystal: the order of summation is a physical choice)Original
The electrostatic energy of an ion in a cubic ionic crystal such as sodium chloride is a sum over all other sites of the lattice. Measuring distances in units of the nearest-neighbour spacing d and putting the reference ion at the origin, the Coulomb energy per ion is −Me2/(4πε0d) with
M=−(i,j,k)=(0,0,0)∑i2+j2+k2(−1)i+j+k,
the Madelung constant, the alternating sign being the alternation of cation and anion along each axis.
This series is not absolutely convergent. The number of lattice sites inside the cube of half-side N is (2N+1)3−1, and their typical distance from the origin is of order N, so the sum of the moduli over that cube grows like a constant times N2: numerically it is 286 at N=5, 3999 at N=20 and 8854 at N=30, close to 9.8N2. The series of moduli therefore diverges, and by the theorems of this unit nothing protects the sum from the order in which it is taken.
And the order matters in practice. Summing over expanding cubes without further care gives the partial values 1.680, 1.701, 1.713, 1.719 at N=8,12,16,20: still more than one per cent from the answer after 69000 sites, because each cube's surface carries a net charge whose potential falls off only slowly. Evjen's remedy is to weight a site on the surface of the cube by the fraction of its charge lying inside, which makes every grouping charge-neutral; the same sites then give 1.747500 at N=5 and 1.7475652 at N=16, agreeing with the accepted value M=1.747565 to six decimal places. Ewald's method achieves the same end differently, by splitting each point charge into a screened short-range part summed in real space and a smooth part summed in reciprocal space.
The mathematics is exactly Riemann's theorem seen from the laboratory: a conditionally convergent lattice sum has no value until a summation scheme is named, and the schemes that work are those whose groupings are neutral, so that the terms being gathered decay fast enough to make the gathered series absolutely convergent. Every molecular-dynamics code that computes Coulomb energies under periodic boundary conditions carries an Ewald sum or a descendant of it for this reason.
Uniform convergence is convergence at a rate independent of the point; it is detected by the Cauchy criterion or the Weierstrass M-test, and it is the hypothesis under which a limit of continuous functions is continuous.
What it means. Pointwise convergence lets each point converge at its own speed, and that is too weak to carry continuity, integrals or derivatives across the limit. Uniform convergence demands one speed for the whole set, which is exactly the strength needed: it is convergence in the supremum distance, so the graphs of fn eventually lie in a narrow band around the graph of f. In practice one proves uniform convergence by bounding the terms of a series by constants and summing the constants, which is the M-test, and one then uses it to justify term-by-term operations.
Let E be a non-empty set and let fn,f:E→C. Both notions below are about the index N beyond which the error is smaller than a prescribed ε>0, and they differ only in whether that index may be chosen after looking at the point.
Pointwise:fn→f pointwise on E when fn(x)→f(x) holds separately at each x∈E; to each ε>0 and each x∈E there then corresponds an index N, which may depend on both, with ∣fn(x)−f(x)∣<ε at every n>N.
Uniform:fn→f uniformly on E when one index serves every point at once: to each ε>0 there corresponds an index N, depending on ε alone, with
∣fn(x)−f(x)∣<εfor all n>N and all x∈E.
The compact form of the second condition is supx∈E∣fn(x)−f(x)∣→0; for E=[a,b] and continuous functions it is convergence in the maximum metric of C[a,b]. Uniform convergence implies pointwise convergence, with the same limit function.
A series ∑n≥1un of functions on Econverges uniformly on E if its partial sums sn=u1+⋯+un converge uniformly on E; writing S for the sum, this says that the tails satisfy supx∈E∣S(x)−sn(x)∣→0. Both notions depend on the set E: a series may converge uniformly on a subset of E and not on E.
Theorem(Cauchy criterion for uniform convergence)AdaptedAdapted from Ahlfors (1979), §2.3, p. 36
Let fn:E→C. Then (fn) has a uniform limit on E exactly when its terms become uniformly close to each other: for each ε>0 some index N, chosen without reference to the point, satisfies
∣fm(x)−fn(x)∣<εwhenever m,n>N and x∈E.
For a series ∑un the same criterion reads ∑n=m+1m+pun(x)<ε for every m≥N, every p≥1 and every x∈E. As in the numerical case, the test names no candidate limit.
Idea of the proof(Cauchy criterion for uniform convergence)Original
One direction is the triangle inequality. For the other, the candidate limit is not given, so it has to be built: at each fixed point the values form a Cauchy sequence of numbers, and the completeness of C supplies their limit. The point of the hypothesis is that the index in it was chosen without reference to the point, so when the limit is taken inside the inequality the index that survives still serves every point at once.
Proof(Cauchy criterion for uniform convergence)AdaptedAdapted from Ahlfors (1979), §2.3, p. 36
Necessity. Suppose fn→f uniformly and let ε>0. Choose N with ∣fn(x)−f(x)∣<ε/2 for all n>N and all x∈E. For m,n>N and any x∈E,
∣fm(x)−fn(x)∣≤∣fm(x)−f(x)∣+∣f(x)−fn(x)∣<ε,
and N depends on ε alone because the two bounds used did.
Sufficiency: build the limit. Suppose the displayed condition holds. Fix x∈E. Reading the condition at that single point says that (fn(x))n is a Cauchy sequence of complex numbers, so by the Cauchy criterion for numerical sequences it converges; define f(x) to be its limit. This defines a function f on E, and it is the only possible uniform limit, since a uniform limit is in particular a pointwise one.
Sufficiency: the convergence is uniform. Let ε>0 and take N from the hypothesis for ε/2, so that ∣fm(x)−fn(x)∣<ε/2 whenever m,n>N and x∈E. Fix such an n and such an x and let m→∞: the left-hand side tends to ∣f(x)−fn(x)∣ by continuity of the modulus, and a non-strict inequality survives the limit, so
∣f(x)−fn(x)∣≤ε/2<εfor all n>N and all x∈E.
The index N was produced by the hypothesis from ε alone, with no reference to x, so this is uniform convergence.
The series form. Apply what has been proved to the partial sums sn, and use sm+p(x)−sm(x)=∑n=m+1m+pun(x) together with the reindexing of pairs (m,p)↔(m,m+p) used for numerical series. □
Let E be a subset of R or of C, let each fn:E→C be continuous at x0∈E, and let fn→f uniformly on E. Then f is continuous at x0. Consequently, if every fn is continuous on E and the convergence is uniform on E, the limit f is continuous on E.
The same holds for a series: if every un is continuous on E and ∑n≥1un converges uniformly on E, its sum is continuous on E.
Idea of the proof(A uniform limit of continuous functions is continuous)Original
Compare f(x) with f(x0) through an intermediary: one approximating function fn, at both points. Two of the three resulting errors are the distance from fn to f, and the third is a continuity estimate for fn. Giving each a third of the budget closes the argument. Uniformity is what lets a single n serve at x0 and at every nearby x, which matters because the nearby points are only identified afterwards, once n has fixed how close they must be.
Proof(A uniform limit of continuous functions is continuous)AdaptedAdapted from Ahlfors (1979), §2.3, p. 36
Let ε>0.
Choose the approximating index first. By uniform convergence there is an n with
∣fn(y)−f(y)∣<ε/3for every y∈E.
Fix that n. This is the step that needs uniformity: the bound must hold at x0 and at every point we may later want to use, and those points are not yet known.
Use continuity of that one function. Since fn is continuous at x0, there is a δ>0 such that every x∈E with ∣x−x0∣<δ satisfies ∣fn(x)−fn(x0)∣<ε/3.
Add the three errors. For such an x, inserting fn(x) and fn(x0) and applying the triangle inequality twice,
the outer terms by step 1 and the middle one by step 2. As ε>0 was arbitrary, f is continuous at x0. Doing this at every x0∈E gives continuity on E.
The series form. Each partial sum sn=u1+⋯+un is a finite sum of functions continuous on E, hence continuous on E, and uniform convergence of the series is uniform convergence of (sn); so parts 1 to 3 apply to the partial sums and give continuity of the sum. □
Let un:E→C and suppose there are constants Mn≥0 with
∣un(x)∣≤Mnfor all x∈E and all n,n≥1∑Mn<∞.
Then ∑n≥1un converges uniformly on E, and it converges absolutely at every point of E. The bounds Mn must not depend on x; that is the whole content of the test.
Idea of the proof(Weierstrass M-test)Original
The numerical series ∑Mn does all the work. Its Cauchy criterion makes every far-out block of Mn small, and the block of un at the same place is smaller in modulus at every point of E. The index at which the blocks become small was computed from the numbers Mn alone, so it cannot depend on the point, and that is uniformity.
The index N came from the numbers Mn, which do not depend on x; the second inequality is where that independence is used. So the series satisfies the Cauchy criterion for uniform convergence, and by the statement proved above it converges uniformly on E.
Absolute convergence at each point. Fix x∈E. Then 0≤∣un(x)∣≤Mn with ∑Mn convergent, so by the comparison test the series ∑n∣un(x)∣ converges; that is absolute convergence at x. □
Remark(Where the hypotheses are used)AdaptedAdapted from Arfken, Weber & Harris (2013), §1.2, Example 1.2.1, pp. 21-22, Example 1.2.2 and Abel's test, p. 23; Ahlfors (1979), §2.3, p. 37
Pointwise is strictly weaker than uniform. On E=[0,1] the functions fn(x)=xn converge pointwise to the function that is 0 on [0,1) and 1 at x=1. Each fn is continuous, the limit is not, so the convergence cannot be uniform; indeed supx∈[0,1]∣fn(x)−f(x)∣=1 for every n. Arfken's Example 1.2.1 makes the same point with the series ∑n≥0(1−x)xn on [0,1], whose sum is 1 on [0,1) and 0 at x=1. Restricting to [0,ρ] with ρ<1 restores uniform convergence, which is why statements about uniform convergence always name the set.
The continuity theorem gives no converse. A pointwise limit of continuous functions can be continuous without the convergence being uniform, so the theorem is a sufficient condition only.
Absolute and uniform convergence are independent. The M-test establishes uniform convergence only for series that also converge absolutely (Ahlfors, p. 37; Arfken, p. 23), so it cannot detect the uniform convergence of ∑n≥1(−1)n/(n+x2) on R, which converges uniformly but not absolutely (Arfken, Example 1.2.2, p. 23). For such series one uses the Cauchy criterion or Abel's test. Arfken states that test on an interval, not on a general set E: if un(x)=anfn(x) where ∑an converges, where fn+1(x)≤fn(x) for all x in [a,b] and all n, and where 0≤fn(x)≤M on [a,b] with M independent of x, then ∑un converges uniformly on [a,b] (Arfken, p. 23). We use it only in that form, on an interval. Conversely a series can converge absolutely at every point without converging uniformly, as the first bullet shows.
Two forms of the M-test. Arfken requires ∣un(x)∣≤Mn for all n (p. 22); Ahlfors asks only for ∣un(x)∣≤Man with a single constant M and for all sufficiently large n, with ∑an convergent (p. 37). The two give the same conclusion, since finitely many terms and a fixed factor affect neither uniformity nor convergence of the comparand.
Of the three tests for uniform convergence named above, two are proved in this unit: the Cauchy criterion and the M-test. Abel's test is only quoted. Arfken does not prove it either, referring for the proof to Knopp and to Whittaker and Watson (p. 23), neither of which is among this handbook's sources. It is used once below, in step 4 of the application, and only in the interval form in which the remark states it.
Example(A series handled completely by the M-test)Original
Consider
S(x)=n≥1∑n2cos(nx),x∈R.
Step 1: uniform convergence. Since ∣cosθ∣≤1 for every real θ,
n2cos(nx)≤n21=Mnfor all x∈R,
and ∑n≥1n−2 is a convergent p-series (p=2>1). The bounds Mn do not involve x, so the M-test applies on the whole of R: the series converges uniformly on R, and absolutely at every point.
Step 2: the sum is continuous. Each term is continuous on R and the convergence is uniform there, so S is continuous on R; it is also 2π-periodic, since every term is.
Step 3: a uniform error bound. For n≥N+1 the function x↦x−2 is at least n−2 on [n−1,n], so n−2≤∫n−1nx−2dx, and summing,
x∈Rsup∣S(x)−sN(x)∣≤n>N∑n21≤∫N∞x2dx=N1.
So N=1000 terms give the value of S to within 10−3 simultaneously at every real x — a statement about the whole function, which is the practical difference uniformity makes.
Step 4: a value to check against. At x=π, cos(nπ)=(−1)n, so S(π)=∑n≥1(−1)n/n2. This series converges absolutely, so its terms may be regrouped freely: splitting ζ(2)=∑n≥1n−2 into even and odd indices gives ∑n evenn−2=41ζ(2), whence
using at the end the closed form ζ(2)=π2/6, which is the case s=2 of the evaluation of the zeta function at even integers (Riemann zeta function and Bernoulli numbers); it is not proved here, and nothing in the rest of the example depends on it. What the example needs is a number, and Arfken's Table 1.1 (p. 17) supplies one independently of the closed form: ζ(2)=1.6449340668, so S(π)=−21ζ(2)=−0.8224670334, which is indeed −π2/12 to the ten digits given. The partial sum s1000(π)=−0.8224665339 differs from it by 5.0×10−7 — inside the bound 1/1000 of step 3, as it must be.
Example(Pointwise convergence with a continuous limit, yet not uniform)Original
On E=[0,1] let
gn(x)=1+n2x2nx.
The pointwise limit is 0. At x=0 every gn(0)=0. For fixed x>0, dropping the 1 in the denominator gives 0≤gn(x)≤nx/(n2x2)=1/(nx)→0.
The supremum does not move. Differentiating,
gn′(x)=(1+n2x2)2n(1−n2x2),
which is positive for x<1/n and negative for x>1/n, so gn increases to a maximum at x=1/n∈[0,1] and then decreases. The maximum value is
gn(n1)=1+11=21for every n.
Hence supx∈[0,1]∣gn(x)−0∣=21 for every n, which does not tend to 0: the convergence is not uniform. The spike keeps its height and merely slides towards the origin, so no single index can bring the error below 21 everywhere at once.
Note what this does and does not show. The limit 0 is continuous, and each gn is continuous, yet the convergence is not uniform — so the continuity theorem above is a sufficient condition and nothing more. Checking supE∣gn−g∣ is the only reliable test; the continuity of the limit is no evidence either way.
Application(Fourier solution of the heat equation)Original
A thin rod occupying 0≤x≤π with both ends held at temperature 0 and initial profile f has temperature
with k>0 the diffusivity. Writing this down is easy; the analytic content is whether the series means anything, and the M-test answers it.
The sets of pairs, read inside C. The theorems above are stated for a set E that is a subset of R or of C, and the sets used here are sets of pairs. Identify the pair (x,t) with the complex number x+it once and for all, so that every such set is a subset of C and the theorems apply verbatim. The identification carries the Euclidean distance of the plane to the modulus, ∣(x+it)−(x0+it0)∣=(x−x0)2+(t−t0)2, and the modulus is all that the continuity theorem and the Cauchy criterion use; continuity of a function of (x,t) in the ordinary plane sense is therefore the same thing as continuity of the corresponding function on a subset of C. Nothing analytic about C — no differentiability in the complex sense — is being invoked.
Uniform convergence away from t=0. The coefficients are bounded: ∣bn∣≤C with C=(2/π)∫0π∣f∣, which is finite for any integrable initial profile, however rough. Fix t0>0 and regard the terms as functions of the pair (x,t) on the set E=[0,π]×[t0,∞), read as a subset of C as just agreed. Then
bne−n2ktsin(nx)≤Ce−n2kt0=Mnon E,
and ∑nMn converges — by the ratio test, since Mn+1/Mn=e−(2n+1)kt0→0. So the series converges uniformly on E, and since each term is continuous in (x,t), the temperature field is continuous there. This is the smoothing property of diffusion in analytic dress: a discontinuous initial profile gives a continuous temperature at every positive time.
Why the set must be named. The constants Mn become C, which is not summable, when t0=0, so the M-test is silent there. It is silent because the conclusion is false, and the failure is worth locating exactly. For this paragraph strengthen the hypothesis on the initial profile: let f satisfy the Dirichlet conditions on [0,π] — at most finitely many discontinuities, all of them finite jumps, at most finitely many maxima and minima, and ∫0π∣f∣<∞ — and let it jump at an interior point x∗, so that the one-sided values f(x∗−) and f(x∗+) exist and differ. Integrability alone, the only assumption used above, does not suffice for what follows: the pointwise convergence theorem quoted in step 1 asks for the Dirichlet conditions, and with mere integrability nothing in this handbook identifies the sum of the sine series at t=0, or even guarantees that it converges there.
What the series does at t=0. Setting t=0 leaves ∑nbnsin(nx), which is the Fourier series of the odd 2π-periodic extension of f; that extension inherits the Dirichlet conditions. Under those conditions the series converges at every point, to the value of the function where it is continuous and to the mean of the two one-sided values at a jump (Riley, Hobson and Bence, §12.1, p. 416 and §12.4, p. 420; see Pointwise convergence of Fourier series). Write σ for its sum. Because f has only finitely many discontinuities, x∗ has points of continuity arbitrarily close on both sides, and at those points σ=f; hence σ(x)→f(x∗−) as x↑x∗ and σ(x)→f(x∗+) as x↓x∗. The two limits differ, so σ is discontinuous at x∗ whatever value it takes there.
Sets that contain a piece of t=0 around the jump. Suppose E′ contains J×{0} for some interval J around x∗. Each truncation uN(x,t)=∑n≤Nbne−n2ktsin(nx) is a finite sum of functions continuous on the whole plane, that is on C under the identification above, so if the series converged uniformly on E′ the continuity theorem would make u continuous on E′, hence continuous in x along J×{0} — where u(x,0)=σ(x), discontinuous at x∗ by step 1. So it does not. One can also see how badly it fails, with no appeal to the theorem: uN(⋅,0) is continuous at x∗, so its values just either side of x∗ are close to the single number uN(x∗,0), while σ is close to f(x∗−) on one side and to f(x∗+) on the other, and one number cannot be nearer than half the gap to both. Hence
x∈Jsupσ(x)−uN(x,0)≥21f(x∗+)−f(x∗−)for every N,
a floor that no number of terms lowers. For the profile f=1 on (π/4,3π/4) and f=0 outside it, whose two jumps have size 1, the floor is 21.
Sets that only approach t=0. The case one actually meets is E′=[0,π]×(0,T], which contains no point with t=0. There u is continuous, so the continuity theorem by itself yields no contradiction; one step bridges the gap. Suppose the series converged uniformly on E′ and let ε>0. By the Cauchy criterion there is an N0 with ∣uM−uN∣<ε/2 on E′ whenever M,N>N0. Each difference uM−uN is continuous on the whole plane and E′ is dense in its closure [0,π]×[0,T], so approaching a point of the closure along points of E′ carries the bound across: ∣uM−uN∣≤ε/2<ε on [0,π]×[0,T]. The truncations are therefore uniformly Cauchy, hence uniformly convergent, on [0,π]×[0,T] — which step 2 forbids, since that set contains [0,π]×{0} and so an interval around x∗ at t=0.
Reaching t=0 is not by itself the obstruction. The closure must contain a segment of t=0across the jump; merely meeting t=0 is not enough. At the end of the rod, on E′={0}×[0,∞), every term vanishes because sin0=0, so the convergence is uniform there with zero error at every N, although the closure meets t=0. The same holds on any single vertical segment {x0}×[0,T], for any x0∈[0,π] and any T>0: put an=bnsin(nx0) and φn(t)=e−n2kt, so that ∑an converges by step 1, φn+1(t)≤φn(t) for every t∈[0,T], and 0≤φn(t)≤1 there — the three hypotheses of Abel's test exactly as the remark above states them, on the interval [0,T] in the role of [a,b] and with φn in the role of its fn. The test then gives uniform convergence in t on [0,T], an interval that includes the endpoint t=0. Non-uniformity is thus a statement about letting x cross the jump while t↓0, not about small t alone.
Uniformity on t≥t0>0, and not on t≥0, is therefore the correct statement, and the deterioration as t0↓0 is why a truncated series needs more and more terms as one looks closer to the initial instant.
A usable error bound. Return to the set E=[0,π]×[t0,∞) and the truncation uN. For n>N we have n2≥n(N+1), so
summing the geometric series at the end. With k=1, t0=0.05 and N=20 this is 4.1×10−10C, against a true tail of 3.0×10−10C: twenty terms suffice at that time, for all x at once. The same argument — not the same bound, since the extra factors change the geometric sum — applied to the series of x- or t-derivatives (whose terms carry extra factors n or n2 and so are still dominated by summable constants), is what licenses differentiating the series term by term and confirming that u solves ut=kuxx — see Integrating and differentiating series term by term.
Uniform convergence on a finite interval lets a limit or a sum be taken through an integral, while term-by-term differentiation needs the differentiated series to converge uniformly; both fail without those hypotheses.
What it means. Integration averages, so it tolerates a small uniform error: a band of width ε around the limit changes the integral by at most ε times the length of the interval. Differentiation reads off slopes, and a function can lie within ε of another while oscillating steeply, so no uniform bound on the functions controls their derivatives. That asymmetry is why the integration theorem assumes uniform convergence of the series itself and the differentiation theorem assumes it of the differentiated series. When you cannot check hypothesis 3, differentiate inside the sum only if the series is a power series, where the differentiated series automatically has the same radius of convergence; see Power series.
Theorem(Term-by-term integration)AdaptedAdapted from Arfken, Weber & Harris (2013), §1.2, p. 24, property 2 and Eq. (1.38)
Let a<b be finite.
Sequence form. Let fn:[a,b]→C be continuous and let fn→f uniformly on [a,b]. Then f is continuous, every integral below exists, and
∫abfn(x)dx⟶∫abf(x)dx,since∫ab(fn−f)≤(b−a)x∈[a,b]sup∣fn(x)−f(x)∣.
Series form. Let every un:[a,b]→C be continuous and let ∑n≥1un converge uniformly on [a,b] to S. Then S is continuous and
∫abS(x)dx=n=1∑∞∫abun(x)dx,
the series on the right being convergent. The same holds with [a,b] replaced by any [α,β]⊆[a,b], uniformly in α and β.
Continuity of the terms is used only to make the integrals exist; the estimate shows that the length of the interval is what converts a uniform bound on the error into a bound on the error of the integral.
Idea of the proofOriginal
Only one estimate is needed. An error that is everywhere smaller than ε can shift an integral over an interval of length b−a by at most ε(b−a), and uniform convergence is exactly the statement that the error is eventually everywhere smaller than any prescribed ε. The series form is then not a second theorem: apply the sequence form to the partial sums, whose integrals are, by linearity of the integral, the partial sums of ∑n∫un.
ProofOriginal
Part 1. Each fn is continuous on [a,b] and the convergence is uniform, so by the continuity of a uniform limit the limit f is continuous on [a,b]. A continuous function on a closed bounded interval is Riemann integrable, so all three integrals ∫abfn, ∫abf and ∫ab(fn−f) exist.
Write δn=supx∈[a,b]∣fn(x)−f(x)∣, which is finite because fn−f is continuous on a closed bounded interval. Monotonicity of the integral applied to the pointwise bound ∣fn−f∣≤δn gives
Uniform convergence says precisely that δn→0, and b−a is a finite constant, so the right-hand side tends to 0 and the integrals converge as claimed. This is the step at which the hypothesis cannot be weakened to pointwise convergence, and the step at which finiteness of b−a is used.
Part 2. Put sN=∑n=1Nun. Each sN is a finite sum of continuous functions, hence continuous, and sN→S uniformly on [a,b] by hypothesis. Part 1, applied to the sequence (sN), gives ∫absN→∫abS, and S is continuous.
By linearity of the integral over a finite sum, ∫absN=∑n=1N∫abun. The left-hand side converges, so the partial sums of the numerical series ∑n∫abun converge; that is what it means for that series to converge, and its sum is limN∫absN=∫abS.
For a subinterval [α,β]⊆[a,b] the same estimate gives ∫αβ(sN−S)≤(β−α)sup[α,β]∣sN−S∣≤(b−a)δN, a bound in which α and β no longer appear; so the convergence holds for every subinterval at once. □
Theorem(Term-by-term differentiation)AdaptedAdapted from Arfken, Weber & Harris (2013), §1.2, p. 24, property 3 and Eq. (1.39)
Let un:[a,b]→C with [a,b] finite, and assume
∑n≥1un converges uniformly on [a,b], with sum S;
each un is differentiable on [a,b] with un′ continuous there;
∑n≥1un′ converges uniformly on [a,b].
Then S is differentiable on [a,b], S′ is continuous, and
S′(x)=n=1∑∞un′(x)for every x∈[a,b].
Hypothesis 3 is the essential one and is not implied by hypotheses 1 and 2: uniform convergence of a series says nothing about the convergence of the differentiated series.
Idea of the proofOriginal
Derivatives are not controlled by uniform bounds, but integrals are, so the proof turns the differentiation claim into an integration claim and uses the theorem just proved. Each partial sum is recovered from its own derivative by the fundamental theorem of calculus; pass to the limit in that identity, which is legitimate because hypothesis 3 lets the integral be taken through the sum, and the limit identity exhibits S as an integral of the continuous function ∑nun′. Differentiating that identity back gives the result.
ProofOriginal
Write g=∑n≥1un′ for the sum of the differentiated series and sN=∑n=1Nun for the partial sums of the original one.
The function g is continuous on [a,b]: each un′ is continuous there by hypothesis 2, and the series ∑nun′ converges uniformly on [a,b] by hypothesis 3, so a uniform limit of continuous functions is continuous applies to it.
Each sN is differentiable on [a,b] with sN′=∑n≤Nun′, a finite sum, and this derivative is continuous by hypothesis 2. The fundamental theorem of calculus therefore gives, for every x∈[a,b],
sN(x)=sN(a)+∫axsN′(t)dt.
Now let N→∞ in that identity, one term at a time. On the left, sN(x)→S(x) by hypothesis 1, and likewise sN(a)→S(a). For the integral, hypothesis 3 gives uniform convergence of ∑nun′ on [a,b], hence on [a,x], and the terms un′ are continuous; so part 2 of the term-by-term integration theorem applies on [a,x] and gives ∫axsN′→∫axg. Hence
S(x)=S(a)+∫axg(t)dt(a≤x≤b).
Since g is continuous on [a,b], its indefinite integral on the right-hand side is differentiable in x with derivative g(x), by the fundamental theorem of calculus once more. Therefore S is differentiable on [a,b] with S′=g=∑n≥1un′, and S′ is continuous because g is. □
Remark(Where the hypotheses are used)Original
Uniform convergence cannot be weakened to pointwise in the integration theorem. On [0,1] put fn(x)=2nxe−nx2. Each fn is continuous, and fn(x)→0 for every x∈[0,1], yet
∫01fn(x)dx=[−e−nx2]01=1−e−n⟶1=0.
The convergence is not uniform: fn attains the value 2ne−1/2 at x=1/2n, so sup[0,1]∣fn∣→∞. The spike carries a fixed amount of area while sliding towards the origin.
The interval must be finite. On [0,∞) put gn(x)=e−x/n/n. Then supx≥0∣gn(x)∣=1/n→0, so gn→0 uniformly, but ∫0∞gn(x)dx=1 for every n. The factor (b−a) in the estimate above is exactly what fails. (A limit theorem that survives on unbounded intervals needs a different hypothesis, dominated convergence rather than uniformity.)
Uniform convergence of a series does not permit differentiation. On R put fn(x)=sin(nx)/n. Then supx∣fn(x)∣=1/n→0, so fn→0 uniformly, while fn′(x)=cos(nx) does not converge at all for most x: at x=π it is (−1)n. To put this in the series form in which the theorem above is stated, telescope: set u1=sinx and un(x)=sin(nx)/n−sin((n−1)x)/(n−1) for n≥2, so that the partial sums of ∑nun are exactly the fn. Then ∑nun converges uniformly on R to 0, satisfying hypotheses 1 and 2, while the partial sums of ∑nun′ are cos(nx), so hypothesis 3 fails and the conclusion fails with it: the sum is the constant 0, whose derivative is 0, while ∑nun′ diverges at x=π. Stakgold and Holst make the same point (p. 112): classically, term-by-term differentiation is available only under restrictive conditions, whereas for distributions it is always allowed, and there sin(nx)→0 in the distributional sense.
Hypothesis 1 can be weakened to convergence at one point. If each un is continuously differentiable on [a,b], ∑un′ converges uniformly on [a,b], and ∑un(x0) converges for a single x0∈[a,b], then ∑un converges uniformly on [a,b] and its sum satisfies S′=∑un′. This is stronger than the form stated by Arfken (p. 24), who assumes uniform convergence of ∑un outright; it follows by integrating the differentiated series from x0 and using the integration theorem above.
A convergent series may not be integrated or differentiated termwise as a matter of course. Two separate warnings in Riley, Hobson and Bence bear on this. Item (v) on p. 131 is the narrower one: the differentiated or integrated series need not have the same convergence properties as the original. The general warning, that termwise operations are not to be performed as a matter of course, is the one in §4.5.2(iii) on p. 135, where power series are then given as the exception.
Part 1 of the integration statement is ours. Arfken's p. 24 states only the series form, property 2. The sequence form, the (b−a)sup estimate and the extension to every subinterval [α,β]⊆[a,b] uniformly in α and β are additions of this handbook, and the proof above establishes them; the series form is then a corollary rather than a separate result.
Example(A geometric cosine series, integrated and differentiated term by term)Original
Take un(x)=2−ncos(nx) for n≥1 and let S(x)=∑n≥1un(x). We shall compute ∫0π/2S and S′(π/2) term by term, and check both against closed forms obtained without any series.
The hypotheses. Each un is continuous on R, and ∣un(x)∣≤2−n for every real x, with ∑n≥12−n=1<∞ (a geometric series, convergent). By the Weierstrass M-test the series converges uniformly on R, so the hypotheses of part 2 of term-by-term integration — continuous terms and a uniformly convergent series — hold on every finite interval. For the derivatives, un′(x)=−n2−nsin(nx) is continuous with ∣un′(x)∣≤n2−n, and ∑n≥1n2−n=2<∞; the M-test again gives uniform convergence, so hypothesis 3 of term-by-term differentiation holds too. Both parts of the integration statement and all three hypotheses of the differentiation statement are therefore available on any [a,b].
The sum in closed form. Summing the geometric series ∑n≥1wn=w/(1−w) at w=eix/2 and taking real parts,
S(x)=Re1−eix/2eix/2=Re2−eixeix=5−4cosx2cosx−1,
the last step on multiplying numerator and denominator by 2−eix=2−e−ix and using ∣2−eix∣2=5−4cosx. At x=π/2 this gives S(π/2)=−1/5.
Integrate term by term. Since ∫0π/2cos(nx)dx=sin(nπ/2)/n, which vanishes for even n and equals (−1)k/(2k+1) at n=2k+1,
The partial sums are 0.5, 0.458333, 0.464583, 0.463467, 0.463684, 0.463640, bracketing the limit from alternate sides.
The same number without series. Because 2cosx−1=−21(5−4cosx)+23, the integrand of step 2 is −21+23(5−4cosx)−1. The substitution t=tan(x/2), under which cosx=(1−t2)/(1+t2) and dx=2dt/(1+t2), turns 5−4cosx into (1+9t2)/(1+t2) and so the second piece into 3∫01dt/(1+9t2)=[arctan3t]01=arctan3, while the first piece contributes −π/4. Hence
∫0π/2S(x)dx=−4π+arctan3=0.4636476090008061.
The series of step 3 is the Maclaurin series of arctan at 1/2 — it is what term-by-term integration of the geometric series 1/(1+t2)=∑k≥0(−1)kt2k over [0,1/2] produces, the M-test with Mk=4−k supplying the uniform convergence there — so its value is arctan21. The two agree exactly: by the addition formula, arctan3−arctan1=arctan1+33−1=arctan21.
Differentiate term by term. By the differentiation theorem, S′(x)=−∑n≥1n2−nsin(nx). At x=π/2 only odd n contribute:
the partial sums being 0.5, 0.125, 0.28125, 0.2265625, 0.244140625, 0.23876953125. Differentiating the closed form of step 2 instead, S′(x)=−6sinx/(5−4cosx)2, so S′(π/2)=−6/25=−0.24 exactly. Both routes give −6/25.
What made it work. The decay 2−n is fast enough that multiplying the terms by n leaves a convergent comparand, which is why hypothesis 3 survived differentiation. Replace 2−n by 1/n2 and the series still converges uniformly (M-test with Mn=1/n2), but the differentiated series −∑nsin(nx)/n does not converge uniformly on any neighbourhood of 0, so hypothesis 3 is lost. Its individual terms are no help here: supx∣sin(nx)/n∣=1/n→0, so the terms do tend to 0 uniformly, and the failure is in the tails. It is visible in the sum, which is (π−x)/2 for 0<x<2π and 0 at x=0, hence discontinuous at 0; a uniformly convergent series of continuous terms has a continuous sum, so this series cannot converge uniformly near 0. Concretely, the tail after N terms, evaluated at x=π/N, is −0.278521, −0.281115, −0.281140 for N=10,100,1000 — it does not shrink, but tends to π/2−Si(π)=−0.2811407.
Reading off Fourier coefficients. A measured signal is often presented already as a uniformly convergent series f(x)=∑ncneinx on [−π,π]. Term-by-term integration against e−imx — legitimate because multiplying a uniformly convergent series by a bounded continuous function leaves it uniformly convergent — collapses the sum by orthogonality to cm=(2π)−1∫−ππf(x)e−imxdx. This is how the coefficients of such an expansion are identified with integrals of the function. The derivation is the one Stakgold and Holst give at p. 128, and this paragraph follows it; that is what the src on this block refers to, the two paragraphs below it being ours.
Verifying a separated-variable solution. The standard solution of the heat equation ut=κuxx on 0≤x≤π with u=0 at both ends is u(x,t)=∑n≥1bne−n2κtsin(nx), with the bn fixed by the initial profile. Checking that this really solves the equation means differentiating the series once in t and twice in x, and that needs hypothesis 3. For t≥t0>0 it holds comfortably: the differentiated terms are bounded by n2∣bn∣e−n2κt0, and ∑nn2e−n2κt0 converges, so the M-test applies and the termwise derivatives are legitimate on x∈[0,π], t≥t0. At t=0 it fails: for a discontinuous initial profile such as a step, bn decays only like 1/n, and ∑nn2bnsin(nx) diverges. The mathematics and the physics agree here — the solution is smooth for every positive time, and no smoothness of the initial data is claimed. The same calculation is what justifies differentiating a Fourier-series solution of the wave or Laplace equation inside the domain while refusing to do so on the boundary.
Where the warning bites. In perturbative calculations one routinely differentiates a series expansion with respect to a parameter — a coupling, a temperature, a frequency. The theorem says that this is safe only when the differentiated series converges uniformly in that parameter on the range of interest. A series whose terms peak more and more sharply as the index grows, which is the mechanism in every counterexample in the remark above, will pass the convergence test and fail the differentiation test.
A power series has a radius of convergence given by Hadamard's formula, inside which it converges absolutely, uniformly on closed subdiscs, and may be differentiated and integrated term by term; at a boundary point where it converges, its sum is the limit of its values along the radius.
What it means. A power series behaves on the inside of its disc exactly as a polynomial does: it converges absolutely, you may differentiate and integrate it as often as you like, and its coefficients are the Taylor coefficients of its sum. The radius of convergence is the only number you need, and Hadamard's formula computes it from the coefficients alone. Everything delicate happens on the boundary circle, where convergence must be examined point by point; Abel's theorem is the one general result there, and it says that convergence at a boundary point is enough to make the sum the limit of the values inside, provided you approach the point without grazing the circle.
Definition(Power series and radius of convergence)AdaptedAdapted from Ahlfors (1979), §2.4, p. 38 and formula (18), p. 38
A power series about the centre z0∈C is a series
n=0∑∞an(z−z0)n,an∈C,
with the convention (z−z0)0=1. Its radius of convergence is the number R∈[0,+∞] defined by Hadamard's formula
R1=n→∞limsup∣an∣1/n,
read with 1/0=+∞ and 1/+∞=0. The set ∣z−z0∣<R is the disc of convergence (for real series, the interval of convergence(x0−R,x0+R)), and ∣z−z0∣=R is the circle of convergence.
Let R be the radius of convergence of ∑n≥0an(z−z0)n.
Inside. The series converges absolutely at every z with ∣z−z0∣<R, and for each ρ with 0≤ρ<R it converges uniformly on the closed disc ∣z−z0∣≤ρ.
Outside. If ∣z−z0∣>R the terms an(z−z0)n are unbounded, so the series diverges.
On the circle. Nothing is asserted at ∣z−z0∣=R: there the series may converge at every point, at none, or at some.
The sum. In ∣z−z0∣<R the sum f is a complex-differentiable (analytic) function of z: at every point z1 of the open disc the difference quotient (f(z)−f(z1))/(z−z1) has a limit as z→z1 in C, the limit being unrestricted in the direction of approach.
In particular the series converges uniformly on every compact subset of the disc of convergence, but in general not on the whole open disc.
Idea of the proofAdaptedAdapted from Ahlfors (1979), §2.4, pp. 38-39
Hadamard's number 1/R is the eventual ceiling of ∣an∣1/n, and a limit superior gives exactly two comparisons, one of them one-sided. Past some index the sequence stays below any number larger than 1/R; and it rises above any number smaller than 1/R for infinitely many n, though not necessarily for all large n. Those two comparisons are the whole proof: inside the disc the first makes the terms eventually dominated by a convergent geometric series, which feeds the Weierstrass M-test and delivers absolute and uniform convergence at one stroke, while outside it the second makes infinitely many terms exceed a geometric sequence with ratio above 1, so the terms are unbounded, which no convergent series permits. The reason the uniform statement needs a radius ρ strictly less than R is that the dominating geometric ratio must be bounded away from 1.
ProofAdaptedAdapted from Ahlfors (1979), §2.4, pp. 38-39
Write w=z−z0, so that the series is ∑n≥0anwn and every assertion concerns ∣w∣; nothing below uses the position of the centre. Put L=limsupn∣an∣1/n, which exists in [0,+∞] by the existence of the limit superior, so that L=1/R under the conventions of the definition.
Part 1, uniform convergence on a closed subdisc. If R=0 there is no ρ with 0≤ρ<R and nothing to prove, so let R>0 and fix such a ρ. Choose ρ1 with ρ<ρ1<R — take ρ1=(ρ+R)/2 if R is finite and ρ1=ρ+1 if R=+∞. Then 1/ρ1>1/R=L, and the limit superior is the least number that the sequence eventually stays below, so there is an index n0 with
∣an∣1/n<ρ11,that is∣an∣<ρ1−n,for all n≥n0.
For ∣w∣≤ρ and n≥n0 this gives ∣anwn∣≤(ρ/ρ1)n=:Mn, a bound that does not depend on w. Since q=ρ/ρ1<1, the series ∑nMn is a convergent geometric series, so the Weierstrass M-test gives uniform and absolute convergence of ∑n≥n0anwn on the closed disc ∣w∣≤ρ. Adding the finitely many terms with n<n0 affects neither uniformity nor absolute convergence, so part 1's uniform assertion holds on ∣w∣≤ρ. This is the step where ρ<R is used: it is what makes q strictly less than 1.
Part 1, absolute convergence at every interior point. Given w with ∣w∣<R, apply the previous paragraph with ρ=∣w∣. So the series converges absolutely at w (the comparison is with a convergent geometric series, and absolute convergence implies convergence by Absolute and conditional convergence), and since w was arbitrary, throughout the open disc.
Part 2, divergence outside. Let ∣w∣>R and choose ρ with R<ρ<∣w∣. Then 1/ρ<1/R=L, and the limit superior is a limit point of the sequence from above, so ∣an∣1/n>1/ρ for infinitely many n. For each such n,
∣anwn∣>(ρ∣w∣)n,
and the base exceeds 1, so these moduli are unbounded. In particular the terms do not tend to 0, and a series whose terms do not tend to 0 diverges (the vanishing-terms criterion).
Part 3. Nothing is asserted on the circle, so there is nothing to prove; the first three series of the remark below show that convergence everywhere, nowhere and almost everywhere on the circle all occur with R=1, so no assertion of this kind could be made.
Part 4. That the sum is complex-differentiable in ∣w∣<R is proved as part of the next statement, whose proof constructs f′ as the sum of the derived series; no further argument is needed here. □
Let f(z)=∑n≥0an(z−z0)n have radius of convergence R>0.
Differentiation. The differentiated series ∑n≥1nan(z−z0)n−1 has the same radius of convergence R, and in ∣z−z0∣<Rf′(z)=n=1∑∞nan(z−z0)n−1.
Repeating this, f has derivatives of every order in the disc, each given by termwise differentiation, and the coefficients are determined by f through
an=n!f(n)(z0),n=0,1,2,…
Integration. For a real series f(x)=∑n≥0an(x−x0)n on (x0−R,x0+R), the series ∑n≥0an(x−x0)n+1/(n+1) also has radius of convergence R and is an antiderivative of f there; equivalently, for x0−R<α≤β<x0+R,
∫αβf(x)dx=n=0∑∞an∫αβ(x−x0)ndx.
In the complex case the same termwise integration is valid along any piecewise-smooth path inside the disc of convergence; this case is not proved in this unit, and the note after the proof says why and where its proof belongs. (Hassani's Thm 10.6.3 states it for any path in the circle of convergence, with no condition on the path.)
Both operations leave R unchanged, but may change the behaviour at the ends of the interval of convergence, which must be tested separately.
Idea of the proofOriginal
Hadamard's formula is insensitive to a polynomial factor in the coefficients, because n1/n→1; that one observation gives the derived and integrated series the same radius as the original. With that in hand the derivative is obtained by a direct estimate rather than by any limit interchange: subtract the derived series from the difference quotient, factorise wn−w0n twice, and what is left is bounded by a constant times ∣w−w0∣, the constant being the value of the twice-derived series at a radius ρ<R. Repeating the argument on the derived series gives all higher derivatives, and setting w=z0 in the k-fold derived series leaves the single term k!ak.
ProofAdaptedAdapted from Ahlfors (1979), §2.4, pp. 39-40
Again write w=z−z0 and f(w)=∑n≥0anwn.
Step 1: n1/n→1. Put n1/n=1+δn with δn≥0. For n≥2 the binomial theorem retains only two of its terms to give n=(1+δn)n≥1+(2n)δn2, whence δn2≤(n−1)/(2n)=2/n. So δn≤2/n→0.
Step 2: the derived and integrated series have radius R. First a remark on limits superior: if bn>0 with bn→1 and cn≥0, then limsupnbncn=limsupncn, because for each ε∈(0,1) we eventually have (1−ε)cn≤bncn≤(1+ε)cn, so the two limits superior differ by a factor between 1−ε and 1+ε for every ε (this also covers the value +∞).
Apply it with cn=∣an∣1/n and bn=n1/n→1 from step 1: the series ∑nnanwn has limsupn(n∣an∣)1/n=limsupn∣an∣1/n=1/R, hence radius R. For w=0 the derived series ∑n≥1nanwn−1 is that series divided by w, so the two converge at exactly the same points, and the derived series has radius R as well. Taking bn=(n+1)−1/n, which also tends to 1 because n1/n≤(n+1)1/n≤21/nn1/n, gives limsupn∣an/(n+1)∣1/n=1/R, so ∑n(an/(n+1))wn has radius R; the integrated series ∑nanwn+1/(n+1) is that one multiplied by w, so again the two converge at the same points and the integrated series has radius R. Iterating the first case, every repeatedly derived series has radius R; in particular ∑n≥2n(n−1)anwn−2 does.
Step 3: f′ is the derived series. Write g(w)=∑n≥1nanwn−1, which converges for ∣w∣<R by step 2. Fix w0 with ∣w0∣<R and choose ρ with ∣w0∣<ρ<R. By step 2 the twice-derived series has radius R>ρ, so by part 1 of the convergence theorem it converges absolutely at ρ; that is,
K:=n≥2∑n(n−1)∣an∣ρn−2<∞.
Now let ∣w∣≤ρ with w=w0. Factorising wn−w0n=(w−w0)∑j=0n−1wjw0n−1−j and subtracting nw0n−1 term by term from that inner sum,
Factorising once more, ∣wj−w0j∣≤jρj−1∣w−w0∣, since the second factor is a sum of j products each of modulus at most ρj−1. With ∣w0n−1−j∣≤ρn−1−j the inner sum is bounded by ρn−2∑j=0n−1j=21n(n−1)ρn−2, so
f(w)−f(w0)−(w−w0)g(w0)≤21K∣w−w0∣2.
Dividing by ∣w−w0∣ shows that the difference quotient differs from g(w0) by at most 21K∣w−w0∣, which tends to 0. Hence f is complex-differentiable at w0 with f′(w0)=g(w0). As w0 was any interior point, f is analytic in ∣w∣<R — which is part 4 of the convergence theorem — and f′ is the derived series. The hypothesis R>0 is used here: with R=0 there is no interior point and no K.
Step 4: all orders, and the coefficients. By step 2 the derived series is itself a power series with radius R, so step 3 applies to it, and by induction f has derivatives of every order in the disc, the k-th being the k-fold derived series
f(k)(w)=n≥k∑n(n−1)⋯(n−k+1)anwn−k.
Put w=0, that is z=z0. Every term with n>k carries a positive power of w and vanishes, and the term n=k is k!ak. So f(k)(z0)=k!ak, which is the coefficient formula.
Step 5: integration on a real interval. Let F(x)=∑n≥0an(x−x0)n+1/(n+1), of radius R by step 2. Step 3 applied to F gives F′=f on (x0−R,x0+R), so F is an antiderivative of f there, and f is continuous because it is differentiable. For x0−R<α≤β<x0+R the fundamental theorem of calculus therefore gives
the middle series being convergent because F is defined at both end points. The same conclusion follows from term-by-term integration, whose hypotheses hold on [α,β] by part 1 of the convergence theorem with any ρ between max(∣α−x0∣,∣β−x0∣) and R.
This proves part 1 in full, the coefficient formula, and part 2 for a real interval. The remaining sentence of part 2, termwise integration along a path in the complex plane, is not proved here; the note below says why and where it belongs. □
Note(The complex-path case of part 2 is deferred)Original
Integration along a path is not available at this point of the handbook. The object ∫γf(z)dz and the bound on its size are set up in contour integrals, whose part 3 is the ML estimate: for h continuous on a piecewise-smooth path γ of length ℓ, ∫γhdz≤ℓsupγ∣h∣. That unit cannot be made a prerequisite of this one, because it depends on this unit in turn: the complex exponential is defined by its power series there, so the dependency runs from power series to contour integrals and a prerequisite in the other direction would be circular. Nothing in this unit's four prerequisites supplies the estimate, and it is not a result of elementary calculus.
Granted the ML estimate, the step is immediate, and this is the argument the later unit's machinery carries out. Let γ be a piecewise-smooth path contained in ∣z−z0∣≤ρ for some ρ<R, of length ℓ, and let sN be the N-th partial sum, a polynomial and hence continuous, as is f by step 3. Part 1 of the convergence theorem gives εN=sup∣z−z0∣≤ρ∣f−sN∣→0, so the estimate applied to h=f−sN yields
Piecewise smoothness is what makes ℓ finite and the integrals defined; it is the condition part 2 states, and Hassani's Thm 10.6.3 (p. 321) states the conclusion for any path in the circle of convergence without naming it. The same bound, in the form "length times sup-norm", is what licenses the termwise integration by which the Taylor series of an analytic function is read off Cauchy's integral formula.
Theorem(Abel's limit theorem)AdaptedAdapted from Ahlfors (1979), §2.5, Thm 3, pp. 41-42
Let ∑n≥0an converge, and put f(z)=∑n≥0anzn, which then has radius of convergence at least 1. Then
f(z)⟶n=0∑∞anas z→1
along any approach for which the quotient ∣1−z∣/(1−∣z∣) stays bounded, that is, within a Stolz angle: a sector of opening less than 180∘ with vertex at 1, symmetric about the segment of the real axis coming in from the left.
Real form (this handbook's corollary, not part of Ahlfors's Thm 3, which is the complex statement at z→1; the last paragraph of the proof below derives it by rescaling). If ∑n≥0anRn converges for some R>0, then
x→R−limn=0∑∞anxn=n=0∑∞anRn,
so the sum of a real power series is continuous on (−R,R] whenever it converges at the right-hand end point, and similarly at the left-hand end point.
Idea of the proofAdaptedAdapted from Ahlfors (1979), §2.5, pp. 41-42
Summing by parts replaces the coefficients by the partial sums sn and produces the representation f(z)=(1−z)∑nsnzn, which displays f(z) as a weighted average of the partial sums with weights (1−z)zn that add to 1. Since sn→s, all but finitely many weights multiply numbers already within ε of s; the finitely many exceptional terms carry the factor 1−z and die as z→1. What the Stolz condition buys is that the total mass of the remaining weights, ∣1−z∣/(1−∣z∣), stays bounded, so the ε is not amplified without limit.
ProofAdaptedAdapted from Ahlfors (1979), §2.5, Thm 3, pp. 41-42
Write sn=a0+⋯+an and s=∑n≥0an=limnsn, which exists by hypothesis.
The radius is at least 1. A convergent series has an→0, so (∣an∣) is bounded, say by C≥1; then ∣an∣1/n≤C1/n→1, so limsupn∣an∣1/n≤1 and R≥1. In particular f is defined on ∣z∣<1.
Summation by parts. With s−1=0 we have an=sn−sn−1, so for every N≥1
on collecting the coefficient of each sn. Fix ∣z∣<1. The sequence (sn) is convergent, hence bounded, so ∑nsnzn converges absolutely, and therefore converges, by Absolute and conditional convergence; also sNzN→0. Letting N→∞,
f(z)=(1−z)n=0∑∞snzn(∣z∣<1).
Subtracting the limit. Since (1−z)∑n≥0zn=1 for ∣z∣<1, we may write s=(1−z)∑n≥0szn and subtract:
f(z)−s=(1−z)n=0∑∞tnzn,tn=sn−s→0.
The estimate. Let ε>0 and choose m with ∣tn∣≤ε for all n≥m. Splitting the sum at m and using ∑n≥m∣z∣n≤1/(1−∣z∣),
∣f(z)−s∣≤∣1−z∣n<m∑∣tn∣+ε1−∣z∣∣1−z∣.
Here the hypothesis enters: on the approach considered, ∣1−z∣/(1−∣z∣)≤K for some constant K, so the second term is at most Kε. The first term is ∣1−z∣ times the fixed finite number Am=∑n<m∣tn∣, so it is below ε once ∣1−z∣<ε/(Am+1). Hence ∣f(z)−s∣≤(K+1)ε for all admissible z close enough to 1, and as ε>0 was arbitrary, f(z)→s.
The two descriptions of the approach agree. If z lies in the sector ∣arg(1−z)∣≤θ<90∘ then Re(1−z)≥cosθ∣1−z∣, and from 1−∣z∣2=2Re(1−z)−∣1−z∣2 and 1−∣z∣≥21(1−∣z∣2) we get 1−∣z∣≥∣1−z∣(cosθ−21∣1−z∣); so once ∣1−z∣≤cosθ the quotient is at most 2/cosθ. Conversely, if the quotient is at most K then Re(1−z)=21(∣1−z∣2+1−∣z∣2)≥21(1−∣z∣)≥∣1−z∣/(2K), so cosarg(1−z)≥1/(2K) and z stays in such a sector.
Real form. Apply the above to the coefficients bn=anRn, whose series converges by hypothesis, and to z=x/R. For real z∈(0,1) the quotient ∣1−z∣/(1−∣z∣) equals 1, so the approach is admissible, and ∑nbnzn=∑nanxn→∑nbn=∑nanRn as x→R−. The left-hand end point is the same statement for bn=an(−R)n. Continuity on (−R,R] follows: inside the interval f is differentiable, hence continuous, and at R the one-sided limit has just been identified with the value. □
Why limsup and not lim. The familiar formula 1/R=limn∣an+1/an∣, stated by Riley, Hobson and Bence as a ratio-test criterion for convergence (§4.5.1, p. 132; the worked example on p. 133 applies it) and by Arfken (p. 29), presumes that this limit exists; Ahlfors sets its agreement with Hadamard's formula in that case as Exercise 7 (p. 41). Take an=1 for even n and an=2 for odd n: then ∣an∣1/n→1, so R=1, while ∣an+1/an∣ alternates between 2 and 1/2 and has no limit. Hadamard's formula needs no hypothesis at all.
Uniform convergence holds on closed subdiscs, not on the open disc. For ∑n≥0zn, with R=1, the tail at z=ρ is ρN/(1−ρ), which is unbounded as ρ→1−; so the convergence is uniform on each ∣z∣≤ρ<1 and not on ∣z∣<1. This is why part 1 is stated with ρ, following Ahlfors (Thm 2(i), p. 38), and it is where this unit's two sources differ: Hassani's Thm 10.6.2 (p. 320) asserts uniform convergence at all points within the circle of convergence, that is on the whole open disc, and the geometric series shows that the stronger form cannot hold. The statement above follows Ahlfors; Hassani's locator is retained on that statement only for part 4, the analyticity of the sum, which the two sources state alike.
The boundary is genuinely undetermined. All three of ∑n≥0zn, ∑n≥1zn/n and ∑n≥1zn/n2 have R=1. The first diverges at every point of ∣z∣=1; the second converges at every such point except z=1; the third converges absolutely at all of them.
Termwise operations can change the end points.∑n≥0xn diverges at x=−1, while its termwise integral ∑n≥0xn+1/(n+1) converges there, to −ln2. Riley, Hobson and Bence state the general warning (p. 135): the interval of convergence is unchanged, the end points must be retested.
Abel's theorem needs the Stolz angle, and has no converse. The restriction ∣1−z∣≤K(1−∣z∣) excludes approaches tangential to the circle. In the other direction, the existence of limx→R−f(x) does not imply that ∑anRn converges: ∑n≥0(−1)nxn=1/(1+x)→1/2 as x→1−, but ∑n≥0(−1)n diverges.
The coefficient formula needs R>0. If R=0 the series defines no function on a neighbourhood of z0 and there is nothing to differentiate. Where R>0, the formula an=f(n)(z0)/n! also yields the uniqueness of the representation: two power series about the same centre that agree on a neighbourhood of it have the same coefficients (Arfken, pp. 30-31).
Example(A radius that only Hadamard's formula gives, then termwise integration)Original
Let
an={2−n,3−n,n even,n odd,f(z)=n≥0∑anzn.
The ratio form is unusable. For even n, ∣an+1/an∣=2n/3n+1→0; for odd n, ∣an+1/an∣=3n/2n+1→∞. The sequence of ratios has no limit, and no finite or infinite value is approached, so neither the ratio form of R nor the ratio test decides anything here.
Hadamard's formula.∣an∣1/n is 1/2 at every even n and 1/3 at every odd n, so its limit superior is 1/2 and R=2.
Independent confirmation, and the circle. Splitting into even and odd indices gives two geometric series,
f(z)=m≥0∑(2z)2m+m≥0∑(3z)2m+1=4−z24+9−z23z,
the first converging exactly for ∣z∣<2 and the second for ∣z∣<3; so the sum converges exactly on ∣z∣<2, confirming R=2. At ∣z∣=2 the even terms have modulus 1, so the terms do not tend to 0 and the series diverges at every point of the circle — one of the three behaviours part 3 of the convergence theorem leaves open. At z=1 the closed form gives f(1)=4/3+3/8=41/24=1.708333, which the partial sums reproduce.
Integrate term by term over [0,1]. On ∣z∣≤1 the M-test applies with Mn=∣an∣≤2−n, so the convergence is uniform there and part 2 of the termwise statement applies on [0,1]:
n=0∑∞n+1an=∫01f(x)dx=∫01(4−x24+9−x23x)dx.
Evaluate the integral in closed form. Partial fractions give 4/(4−x2)=1/(2−x)+1/(2+x), whose integral is ln2−x2+x, equal to ln3 at x=1; and ∫3xdx/(9−x2)=−23ln(9−x2), contributing 23ln89. Hence
Check. The partial sums of ∑nan/(n+1) are 1, 1.166667, 1.25, 1.259259, 1.271759, 1.272445, 1.274677, 1.274734, and by n=11 they have reached 1.275263, climbing to the stated value. The integrated series has radius 2 as well, in accordance with step 2 of the proof.
Example(Abel's theorem at a boundary point)Original
We evaluate ∑n≥0(−1)n/(3n+1)=1−41+71−101+⋯, a conditionally convergent series, by working inside the disc of convergence and transporting the answer to the boundary.
A power series whose coefficients are these numbers. For ∣t∣<1 the geometric series with ratio −t3 gives 1/(1+t3)=∑n≥0(−1)nt3n. On [−ρ,ρ] with ρ<1 we have ∣(−1)nt3n∣≤ρ3n, so the M-test applies with Mn=ρ3n and the convergence is uniform on that closed interval. Given x with ∣x∣<1, choose ρ with ∣x∣≤ρ<1; then [0,x] (or [x,0] for negative x) lies in [−ρ,ρ], and term-by-term integration from 0 to x gives
g(x):=n=0∑∞3n+1(−1)nx3n+1=∫0x1+t3dt.
Only the exponents 3n+1 carry non-zero coefficients, and ∣1/(3n+1)∣1/(3n+1)→1, so R=1.
Convergence at the end point. At x=1 the series is alternating with terms 1/(3n+1) decreasing to 0, so it converges by the alternating-series test. It does not converge absolutely, so the M-test gives nothing on the closed interval and the value at x=1 is not available by a uniform-convergence argument.
Abel's limit theorem. Its real form with R=1 now applies and identifies the end-point sum with the limit from inside:
n=0∑∞3n+1(−1)n=x→1−limg(x).
Compute the limit. The function x↦∫0xdt/(1+t3) is continuous at x=1, the integrand being continuous on [0,1], so the limit is ∫01dt/(1+t3). With 1+t3=(1+t)(t2−t+1),
1+t31=3(1+t)1+3(t2−t+1)2−t,
The first piece integrates to 31ln2. For the second, writing 2−t=−21(2t−1)+23 splits it as −61∫01(2t−1)dt/(t2−t+1)+21∫01dt/(t2−t+1). The first of these vanishes, since ∫01(2t−1)dt/(t2−t+1)=[ln(t2−t+1)]01=ln1−ln1=0; the second is 21∫01dt/((t−21)2+43)=21⋅32[arctan32t−1]01=31⋅3π=33π. Hence
Check, and why Abel is needed. The partial sums are 1, 0.75, 0.892857, 0.792857, 0.869780, 0.807280, 0.859912, 0.814457 — they straddle 0.8356488 and converge slowly, as a conditionally convergent series must. Summing enough terms to confirm the four-figure value directly would take thousands of them; Abel's theorem replaces that with an integral evaluated in closed form. The theorem is also not a formality here: the boundary sum and the radial limit g(1−) are two different numbers until a theorem identifies them, and the implication runs only one way. The fifth bullet of the remark above gives the one-way example: ∑n≥0(−1)nxn has the radial limit 21 at x=1 while ∑n≥0(−1)n diverges, so a radial limit can exist with no boundary sum to match it, and no converse of Abel's theorem could do the work done here.
ApplicationOriginal
Series solutions of differential equations. Substituting y=∑n≥0anxn into a linear differential equation and differentiating term by term — legitimate inside the disc of convergence, by the statement above, and the reason the manipulation is allowed at all — converts the equation into a recursion for the coefficients. For Airy's equation y′′=xy, matching powers gives (n+2)(n+3)an+3=an, so ∣an∣ falls off faster than any geometric sequence, limsup∣an∣1/n=0, and the radius of convergence is infinite: the two Airy functions are represented by their Maclaurin series at every real and complex x. That is what makes the series usable on both sides of a linear turning point in the one-dimensional Schrödinger equation, where the classically allowed and forbidden regions must be joined. By contrast, for Legendre's equation the recursion gives R=1, the distance from the origin to the equation's singular points x=±1; the series then converges on −1<x<1 and nothing in this unit settles its behaviour at the poles of the sphere, which is precisely where the physical requirement of boundedness forces the series to terminate and quantises the degree.
Reading a physical singularity off the coefficients. A perturbation expansion in a coupling g is a power series, and Hadamard's formula turns its high-order coefficients into the radius of convergence, which is the distance from g=0 to the nearest singularity of the quantity being expanded. Computing ∣an∣1/n for the available orders is the standard diagnostic: a limit of 0 indicates an entire function of the coupling, a positive limit locates a critical coupling, and coefficients growing like n! indicate a zero radius and an expansion that is asymptotic rather than convergent. In statistical mechanics the same reading applies to the grand partition function as a power series in the fugacity, whose radius of convergence marks the onset of condensation.
Summing at the edge. Abel's limit theorem is the licence to evaluate a generating function or lattice sum inside its disc, in closed form, and then let the variable approach a boundary point at which the series converges, as in the second example above. The Stolz-angle restriction matters in the complex case: approaching a boundary point along the circle itself, or tangentially to it, is not covered, which is why numerical analytic continuation to a boundary point is done along a radius or inside a sector.
The Taylor polynomial of order N approximates a function with an error given exactly by the Lagrange or the integral remainder; the Taylor series represents the function only where that remainder tends to zero, and for some smooth functions it never does.
What it means. Taylor's theorem is not a statement about infinite series: for each fixed N it is an exact identity, a polynomial plus an error term you can bound. That is what makes it the standard tool for approximation and for error estimates, and the Lagrange form is usually the one to reach for, because it converts a bound on one derivative into a bound on the error. The infinite Taylor series is a separate question, and the right one to ask is whether the remainder tends to zero, not whether the series converges: it can converge to the wrong thing, or converge nowhere useful, while the function remains perfectly smooth.
Let f be real-valued on an interval I, let a∈I, and let N≥0 be an integer such that f(N)(a) exists. The Taylor polynomial of order N of f about a is
PN(x)=k=0∑Nk!f(k)(a)(x−a)k,
and the remainder is RN(x)=f(x)−PN(x). With a=0 the polynomial is called a Maclaurin polynomial. If f has derivatives of all orders at a, the Taylor series of f about a is ∑k≥0f(k)(a)(x−a)k/k!; it is a power series in x−a, so Power series applies to it. The series represents f at a point x exactly when RN(x)→0 as N→∞; convergence of the series alone is not enough, because convergence of PN(x) to some limit does not identify that limit as f(x), and the two can differ at every x of an interval (first bullet of the remark below).
Let N≥0, let a and x be distinct points of I, let J be the closed interval with end points a and x (so J=[a,x] if x>a and J=[x,a] if x<a), and suppose f(N+1) exists and is continuous on J. Then
f(x)=k=0∑Nk!f(k)(a)(x−a)k+RN(x)
with the remainder given in either of the two exact forms
RN(x)=(N+1)!f(N+1)(ξ)(x−a)N+1for some ξ between a and x,RN(x)=N!1∫ax(x−t)Nf(N+1)(t)dt.
The first is the Lagrange form, the second the integral form. Both are identities, not approximations: no terms are discarded. The Lagrange form gives the error bound
∣RN(x)∣≤(N+1)!∣x−a∣N+1t∈Jmaxf(N+1)(t),
and if f has derivatives of all orders on an interval I containing a and this bound tends to 0 as N→∞ for every x∈I, then the Taylor series converges to f on I.
The stated hypothesis, continuity of f(N+1) on the closed interval J, is the one Arfken assumes, and it serves both remainder forms. The integral form needs it, since the integrand must be integrable on J; the Lagrange form alone holds under less, and the sharp hypothesis for it is given in the remark below.
Both remainder formulas come from one identity. Freeze x and slide the expansion point: the function t↦∑k≤Nf(k)(t)(x−t)k/k! equals the Taylor polynomial at t=a and equals f(x) at t=x, and when it is differentiated almost everything cancels, leaving a single term. Integrating that one term from a to x gives the integral form of the remainder; replacing the integral by its weighted mean value gives the Lagrange form, and the error bound is the crudest estimate of that mean.
The term of the first sum with index k equals the term of the second with index k+1, so these cancel in pairs and only the last term of the first sum survives:
Φ′(t)=N!f(N+1)(t)(x−t)N.
By hypothesis f(N+1) is continuous on J, so Φ′ is continuous there and the fundamental theorem of calculus gives Φ(x)−Φ(a)=∫axΦ′(t)dt. Now Φ(x)=f(x), since every term with k≥1 carries a factor (x−x)k=0, and Φ(a)=PN(x) by the definition of the Taylor polynomial. Hence
RN(x)=f(x)−PN(x)=N!1∫ax(x−t)Nf(N+1)(t)dt,
the integral being the oriented one, so that the formula covers x>a and x<a alike.
A lemma: the weighted mean value theorem for integrals.Let h and w be continuous on a closed bounded interval [c,d] with w≥0 there. Then ∫cdhw=h(ξ)∫cdw for some ξ∈[c,d]. For h attains a least value m and a greatest value M on [c,d], being continuous on a closed bounded interval; since w≥0 this gives mw≤hw≤Mw pointwise, and integrating, m∫cdw≤∫cdhw≤M∫cdw. If ∫cdw=0 then both sides of the claim are 0 and any ξ serves. Otherwise (∫cdhw)/(∫cdw) lies in [m,M], and the intermediate value theorem supplies a ξ∈[c,d] at which h takes that value. This is the weighted form of the mean value theorem that Arfken uses at his Eq. (1.42); the weight is what lets it be applied once instead of N times.
The Lagrange form for x>a. Apply the lemma on [a,x] with h=f(N+1), continuous by hypothesis, and w(t)=(x−t)N/N!, which is non-negative there. Since ∫ax(x−t)Ndt/N!=(x−a)N+1/(N+1)!,
RN(x)=f(N+1)(ξ)(N+1)!(x−a)N+1,ξ∈[a,x].
The Lagrange form for x<a. Reverse the orientation: RN(x)=−N!1∫xa(x−t)Nf(N+1)(t)dt, and on [x,a] we have (x−t)N=(−1)N(t−x)N with (t−x)N≥0. The lemma with weight w(t)=(t−x)N/N!, whose integral over [x,a] is (a−x)N+1/(N+1)!, gives RN(x)=(−1)N+1f(N+1)(ξ)(a−x)N+1/(N+1)!; and (−1)N+1(a−x)N+1=(x−a)N+1, so this is the same formula, with ξ∈[x,a]. In both cases ξ lies between a and x.
The error bound. Take moduli in the Lagrange form. Since f(N+1) is continuous on the closed bounded interval J, the maximum maxt∈J∣f(N+1)(t)∣ exists and dominates ∣f(N+1)(ξ)∣ wherever ξ happens to lie.
Convergence of the Taylor series. Fix x∈I. If the bound tends to 0 as N→∞, then so does RN(x), that is PN(x)→f(x); and PN(x) is the Nth partial sum of the Taylor series at x. If this holds at every x∈I, the series converges to f throughout I. □
Theorem(The CN form of the remainder)AdaptedAdapted from Folland (1999), §8.1, p. 236 and §8.8, p. 278
Let f be N times continuously differentiable on an open interval containing a. Then
x→alim(x−a)NRN(x)=0,
that is, RN(x)=o(∣x−a∣N) as x→a. Here only N derivatives are assumed, not N+1, and no point ξ and no integral appear; the conclusion is correspondingly weaker, being a statement about the limit x→a and not a bound at a fixed x.
Idea of the proofOriginal
Use the previous theorem one order lower. At order N−1 the Lagrange remainder needs only N derivatives, which is all that is available, and it expresses the error as f(N) evaluated at an unknown intermediate point. The difference between that and the order-N Taylor polynomial is the single term f(N)(a)(x−a)N/N!, so the whole remainder is (f(N)(ξ)−f(N)(a))(x−a)N/N!, and continuity of f(N) at a makes the bracket vanish in the limit.
ProofOriginal
For N=0 we have P0(x)=f(a), so the assertion reads f(x)−f(a)→0 as x→a: that is the continuity of f, which the hypothesis f∈C0 supplies. So let N≥1, and let U be an open interval containing a on which f is N times continuously differentiable.
Fix x∈U with x=a and let Jx⊂U be the closed interval with end points a and x. Apply the previous theorem on Jx at order N−1: its hypothesis is that f((N−1)+1)=f(N) exists and is continuous on Jx, which the CN hypothesis supplies — and note that the (N+1)st derivative is never mentioned. The Lagrange form gives a point ξx between a and x with
f(x)=k=0∑N−1k!f(k)(a)(x−a)k+N!f(N)(ξx)(x−a)N.
Subtract PN(x)=∑k=0Nf(k)(a)(x−a)k/k!. The two polynomial sums differ by exactly the term k=N, so
RN(x)=f(x)−PN(x)=N!f(N)(ξx)−f(N)(a)(x−a)N.
Divide by (x−a)N, which is non-zero because x=a:
(x−a)NRN(x)=N!f(N)(ξx)−f(N)(a).
Finally let x→a. Since ξx lies between a and x we have ∣ξx−a∣≤∣x−a∣→0, so ξx→a; and f(N) is continuous at a, which is the hypothesis used here and nowhere else, so the numerator tends to 0. Hence the quotient tends to 0, which is the assertion. □
Each series below converges to the stated function at every real x in the stated range, and diverges at every real x outside the closure of that range, with the one exception noted after the table.
Function
Maclaurin series
Range
ex
∑n≥0xn/n!
all x
sinx
∑n≥0(−1)nx2n+1/(2n+1)!
all x
cosx
∑n≥0(−1)nx2n/(2n)!
all x
ln(1+x)
∑n≥1(−1)n−1xn/n
−1<x≤1
arctanx
∑n≥0(−1)nx2n+1/(2n+1)
−1≤x≤1 (ours)
(1+x)α
∑n≥0(nα)xn
∣x∣<1 (ours)
The two ranges marked "(ours)" are not the ranges printed by either cited source: Riley, Hobson and Bence give −1<x<1 for the arctangent and −∞<x<∞ for the binomial series (p. 140). Those two rows rest on the argument in the last bullet of the remark below and on the proof below, not on the sources; the other four ranges are the sources'.
In the last line α∈R and (nα)=α(α−1)⋯(α−n+1)/n!, with (0α)=1. If α is a non-negative integer the series has only finitely many non-zero terms and equals (1+x)α for every real x; otherwise its radius of convergence is exactly 1, and whether it converges at x=±1 depends on α.
Idea of the proofOriginal
The six rows divide into three kinds. For ex, sin and cos every derivative is bounded on each bounded interval by a constant that does not grow with the order, so the Lagrange bound has a factorial in the denominator and nothing to fight it: the series converges to the function everywhere. For the logarithm and the arctangent the remainder is awkward but the function is an integral of a geometric series, so termwise integration delivers the series at a stroke on ∣x∣<1, and Abel's limit theorem then carries it to the end points at which it converges. The binomial series is proved to sum to (1+x)α not by its remainder but by showing that its sum solves the same first-order differential equation as (1+x)α, with the same value at x=0.
ProofOriginal
Throughout, a=0 and J is the closed interval with end points 0 and x. All six series are power series in x, so Power series governs their convergence, and in particular a series of radius 1 has unbounded terms, hence diverges, at every ∣x∣>1. That disposes of the "diverges outside the closure" half of the claim in the last three rows; the first three rows have no outside.
ex. Here f(N+1)(t)=et, so maxt∈J∣f(N+1)(t)∣=emax(0,x)≤e∣x∣, and the error bound of Taylor's theorem is ∣RN(x)∣≤∣x∣N+1e∣x∣/(N+1)!. For each fixed x this tends to 0, because ∣x∣N+1/(N+1)!→0. So the series converges to ex at every real x. Its coefficients are f(k)(0)/k!=1/k!.
sinx and cosx. Every derivative of either function is ±sin or ±cos, so maxJ∣f(N+1)∣≤1 and ∣RN(x)∣≤∣x∣N+1/(N+1)!→0 for each fixed x. The coefficients follow from the derivative values at 0, which run 0,1,0,−1 for sin and 1,0,−1,0 for cos, giving the two series as displayed.
ln(1+x). For ∣t∣<1 the geometric series with ratio −t gives 1/(1+t)=∑n≥0(−1)ntn, of radius 1. By termwise integration of a power series, integrating from 0 to x for ∣x∣<1 gives ln(1+x)=∑n≥1(−1)n−1xn/n, which is therefore the Maclaurin series and converges to the function on (−1,1). At x=1 the series is alternating with terms 1/n decreasing to 0, so it converges by the alternating-series test, and Abel's limit theorem identifies its sum with limx→1−ln(1+x)=ln2. At x=−1 the series is −∑n1/n, the harmonic series, which diverges. The range is thus −1<x≤1.
arctanx. The same route with ratio −t2: 1/(1+t2)=∑n≥0(−1)nt2n of radius 1, and termwise integration gives arctanx=∑n≥0(−1)nx2n+1/(2n+1) for ∣x∣<1. At x=1 and at x=−1 the series is alternating with terms 1/(2n+1) decreasing to 0, so it converges at both; Abel's limit theorem, applied at each end point, identifies the sums with arctan(±1)=±π/4. The range is therefore the closed interval −1≤x≤1, which is the entry marked "(ours)" in the table.
(1+x)α. If α is a non-negative integer then (nα)=0 for n>α, the series is the finite binomial expansion of a polynomial, and it equals (1+x)α at every real x. Otherwise no coefficient vanishes and
(nα)(n+1α)=n+1∣α−n∣⟶1,
so the radius of convergence is exactly 1 and the series diverges for ∣x∣>1. Let g(x)=∑n≥0(nα)xn on (−1,1). From (n+1α)=(nα)(α−n)/(n+1), termwise differentiation (legitimate inside the radius) gives
the second by re-indexing the first form of g′. Adding the two displays, the terms in n cancel and
(1+x)g′(x)=αn≥0∑(nα)xn=αg(x)(∣x∣<1).
Hence dxd[(1+x)−αg(x)]=(1+x)−α−1[(1+x)g′(x)−αg(x)]=0 on (−1,1), so (1+x)−αg(x) is constant there; at x=0 it equals g(0)=1. Therefore g(x)=(1+x)α on ∣x∣<1, which is the entry marked "(ours)". □
A smooth function need not be the sum of its Taylor series. Let η(x)=e−1/x for x>0 and η(x)=0 for x≤0. Then η is infinitely differentiable on all of R, including at the origin, where every derivative vanishes (Folland, p. 236, builds the standard bump functions from exactly this η). Its Maclaurin series is therefore identically 0, which converges everywhere but equals η only on (−∞,0]. So "the Taylor series converges" and "the Taylor series represents f" are different statements, and the remainder is what separates them: here RN(x)=η(x)=0 for every N and every x>0.
Hypotheses actually needed. The two remainder formulas use continuity of f(N+1) on the closed interval, as Arfken assumes (p. 26). For the Lagrange form alone, less suffices: it is enough that f(N) be continuous on J and f(N+1) exist at every interior point, and then ξ may be taken strictly between a and x. Arfken notes in footnote 2 (p. 26) that the expansion holds under slightly less restrictive conditions than he assumes, and Folland remarks (p. 278) that calculus texts usually require N+1 derivatives although the CN statement above does not. Riley, Hobson and Bence give the remainder formula without stating any differentiability hypothesis (pp. 137-138); the one above is the standard hypothesis under which their formula holds.
Indexing and other remainder forms. Arfken's Eq. (1.40) is indexed by the number of terms: his Rn with polynomial degree n−1 is our RN with N=n−1. His Eq. (1.41) gives the remainder as an n-fold iterated integral of f(n); carrying out all but one of the integrations turns it into the single integral above, which is the one-variable case of the Banach-space formula of Stakgold and Holst (Thm 10.1.2, p. 647). Arfken's footnote 3 records a third version, due to Cauchy. Folland's indexing differs again: his k is our N, and his statement on p. 236 is the multi-variable one, a sum over multi-indices ∣α∣≤k on Rn with remainder o(∣x−x0∣k); the CN statement above is its specialisation to n=1, where the multi-index sum becomes the ordinary Taylor polynomial of order N.
The ranges in the table, compared with the sources. Riley, Hobson and Bence print −1<x<1 for arctanx and −∞<x<∞ for (1+x)n (p. 140). At x=±1 the arctangent series converges by the alternating-series test, and Abel's limit theorem identifies its sum with arctan(±1)=±π/4, which is why the closed range is given above. The unrestricted range for the binomial series is the correct one in the case their notation suggests, a non-negative integer exponent n, for which the series terminates; for an exponent that is not a non-negative integer every coefficient is non-zero and (n+1α)/(nα)=∣α−n∣/(n+1)→1, so the radius of convergence is 1 and the range in the table above is ∣x∣<1. The logarithmic range −1<x≤1 agrees with both sources (Arfken, p. 29; Riley, Hobson and Bence, p. 140); at x=1 it is the alternating harmonic series, with sum ln2, and at x=−1 the harmonic series, which diverges.
Example(How good is the bound, and how many terms does one need)Original
Approximate f(x)=(1+x)−1/2 near 0 and compare the predicted error with the true one. This is the binomial row of the table with α=−21, so the series converges to f on ∣x∣<1; the question here is quantitative, at the fixed point x=0.3.
The derivatives. Differentiating k times, f(k)(x)=(−1)k4kk!(2k)!(1+x)−1/2−k, which gives f(k)(0)=(−1)k(2k)!/(4kk!): the values 1,−21,43,−815,16105,−32945 for k=0,…,5. Note that ∣f(k)(t)∣ decreases in t, so on [0,0.3] its maximum is at t=0.
The order-4 polynomial. Dividing by k!,
P4(x)=1−2x+83x2−165x3+12835x4.
The true error at x=0.3.P4(0.3)=0.877527343750000 and f(0.3)=1/1.3=0.877058019307029, so
R4(0.3)=f(0.3)−P4(0.3)=−4.693244×10−4.
The predicted bound. With N=4, J=[0,0.3] and maxJ∣f(5)∣=∣f(5)(0)∣=945/32=29.53125, the error bound of Taylor's theorem gives
The bound holds, and it overstates the true error by only 27%: the ratio of true error to bound is 0.785.
Where ξ actually is. The Lagrange form says R4(0.3)=f(5)(ξ)(0.3)5/5! for some ξ∈(0,0.3). Solving −(945/32)(1+ξ)−11/2=120R4(0.3)/0.35=−23.1765 gives ξ=0.04504, comfortably inside the interval, and much closer to 0 than to 0.3 — which is why the bound, computed at the worst point t=0, is nearly attained.
How many terms for eight-figure accuracy at x=0.3? The bound at order N is 0.3N+1∣f(N+1)(0)∣/(N+1)!. It takes the values 5.98×10−4 at N=4, 3.65×10−6 at N=8, 2.47×10−8 at N=12 and 7.15×10−9 at N=13. So N=13 is the first order at which the theorem guarantees ∣RN(0.3)∣<10−8. The true errors are 1.92×10−8 at N=12 and 5.54×10−9 at N=13, so N=13 is also the first order that genuinely achieves it: at this point the bound costs nothing. This is the practical content of the Lagrange form — a bound on one derivative is converted into a usable number of terms, without evaluating the function being approximated.
ApplicationOriginal
Truncation error of a finite-difference scheme. Discretising a second derivative on a grid of spacing h uses the centred second difference. Taylor's theorem with the Lagrange remainder at order 3, applied at x+h and at x−h for a function u with u(4) continuous near x, gives u(x±h)=u(x)±hu′(x)+21h2u′′(x)±61h3u′′′(x)+241h4u(4)(ξ±) with ξ± between x and x±h. The odd terms cancel on adding, so
This is the second-order accuracy of the discrete Laplacian used in every finite-difference solver for Poisson's and the heat equation, and the bound is what fixes the grid spacing needed for a target accuracy. For u=sin at x=1 with h=0.1: the difference quotient is −0.840769992687 against u′′(1)=−0.841470984808, an error of 7.0099×10−4; the bound is h2sin(1.1)/12=7.4267×10−4 and the leading term h2u(4)(1)/12=h2sin(1)/12=7.0123×10−4, which matches the true error to three figures. Halving h divides the error by four, as the h2 predicts.
Harmonic approximation and its range of validity. Expanding a potential V about a minimum q0, where V′(q0)=0, Taylor's theorem gives V(q)=V(q0)+21V′′(q0)(q−q0)2+61V′′′(ξ)(q−q0)3. The harmonic oscillator with frequency V′′(q0)/m is what remains when the last term is dropped, and the remainder is exactly the anharmonicity: the frequency is reliable only while ∣V′′′∣∣q−q0∣3/6 is small against 21V′′(q0)(q−q0)2, that is while ∣q−q0∣≪3V′′(q0)/max∣V′′′∣. For the pendulum, V(θ)=mgl(1−cosθ) has V′′′(0)=0, so the first correction is quartic and the harmonic range is unusually wide — which is why the simple pendulum formula survives to amplitudes of tens of degrees.
Non-relativistic limits. The binomial row with α=−21 turns the relativistic energy mc2(1−v2/c2)−1/2 into mc2+21mv2+83mv4/c2+⋯, and the remainder bound says how fast the vehicle may move before the Newtonian kinetic energy is inadequate at a prescribed tolerance. The same expansion with the same bound is what justifies the dipole and quadrupole truncations of a multipole expansion, where ∣x∣ is the ratio of source size to distance.