WHAT IS... ABEL SUMMATION?!

HADRIAN QUAN

Abstract. This is an expository note for myself and the members my REU team, with the goal of un- derstanding what Jean-Paul Allouche meant in his paper “Paperfolding Infinite Products and the Gamma Function” when he says “This is an immediate consequence of Abel Summation”. This note is meant to be self-contained. In single variable calculus the integration by parts formula, Z b Z b f(x)0g0(x)dx = f(b)g0(b) − f(a)g0(a) − f(x)g00(x)dx a a is a wonderful trick to keep in a students toolbox. It can help with the evaluation of particularly difficult integrals. Can a similar technique be used to simplify infinite series? The insight comes from the analogy that sums are like integrals, and differences are like derivatives. With this in mind we prove a lemma of Abel Pn Pn Lemma 0.1 (Abel Summation By Parts). Given two finite sums k=1 ak and k=1 bk, define An = Pn k=1 ak. Then we have n n X X akbk = Ak(bk − bk−1) + Anbn − Am−1bm k=m k=m

Proof. First note that from these definitions it follows that am = Am − Am−1. Then the proof involves the following computation, which amounts to rearranging the terms in this sum: n n n n−1 n n−1 X X X X X X akbk = (Ak − Ak−1)bk = Akbk − Ak−1bk = Akbk − Akbk+1 k=m k=m k=m k=m k=m k=m−1 n−1 n−1 X X = Akbk + Anbn − Am−1bm − Akbk+1 k=m k=m n−1 X = Ak(bk − bk−1) + Anbn − Am−1bm k=m as claimed.  Now that we have this lovely summation by parts formula, we can get to using it, to prove some results! But first, more definitions! ∞ n P P Definition 0.2 (Cauchy Criterion). The series xk = lim xk = lim sn converges if and only if k=1 n→∞ k=1 n→∞ 1 ∀ r > 0 small, ∃m ∈ N such that j X 1 x = |s − s | ≤ k j i r k=i for all i, j ≥ m The idea behind this convergence criterion is, this condition makes the limit of partial sums into a Cauchy sequence, and hence the sequence converges to some real number. Intuitively, if the value of the sum gets small for the sum of terms far enough out in the sequence, this guarantees convergence of the sum. Now lets prove the formula for Abel Summation. This will give us a stronger form of the “Alternating Series Test” from calc 2.

Date: December 2014. 1 2 HADRIAN QUAN

Theorem 0.3 (Abel Summation). Given two sequences of real numbers {ak} and {bk} which satisfy Pn • The partial sums An = k=1 ak form a bounded sequence • The sequence {bk} is monotone decreasing • and limk→∞ bk = 0 P∞ then the series k=1 akbk converges. 1 Proof. To show convergence we will show this series satisfies the Cauchy Criterion. Let r > 0 be given. Let |An| ≤ M < +∞ for all n, since the An are bounded. Since limk→∞ bk = 0, then ∃m ∈ N such that 1 bN ≤ 2Mr . Then for all n, m ≥ N we have n n−1 n−1 X X X 1 a b = A (b − b ) + A b − A b ≤ M (b − b ) + b − b = 2Mb ≤ 2Mb ≤ k k k k k+1 n n m−1 m k k+1 n m m N r k=m k=m k=m hence this series satisfies the Cauchy Criterion and so it converges.  k Note that for the case of the alternating series test, set ak = (−1) . Now, lets prove convergence of the series considered by Allouche. In his paper, in proposition 3, Allouche has that {k} is a sequence Pn for which the growth of the partial sums is bounded, i.e. ∃M ∈ R such that k=1 k ≤ Mlog(n) for all n ≥ 0. Additionally f : R → C such that f(x) is monotone decreasing to zero as x → ∞. Then to prove P∞ k=1 kf(k) is convergent, we use Abel Summation! Since bk → 0 then ∀n ∈ N there exists l ∈ N such that 1 bm ≤ n for all m ≥ l. Hence

n n−1 n−1 X X X kf(k) = Ek(bk − bk+1) + Enbn − Em−1bm ≤ Mlog(n) (bk − bk+1) + bn − bm k=m k=m k=m

2Mlog(n) 1 = 2Mlog(n)b ≤ = 2Mlog(n n ) m n 1 1 and as n → ∞, n n → 1, so log(n n ) → 0. Hence this series converges.