Abel's Test

Abel’s Test | Abel’s Test, Uniform Version

This note is an exposition of Abel’s test on convergence of series.

Theorem 1. Suppose $\sum _1^{\mathrm{\infty }}{b}_n$ converges and that {a_n} is a monotone bounded sequence. Then $\sum _1^{\mathrm{\infty }}{a}_n{b}_n$ converges.

Proof. Let b₀ = 0, $B_N=\sum _{k=0}^N{b}_k$. Then b_n = B_n − B_n − 1, n ≥ 1, hence $$\begin{array}{rl}\sum _{k=1}^N{a}_k{b}_k& =\sum _{k=1}^N{a}_k(B_k-B_{k-1})\\ & =B_1(a_1-a_2)+B_2(a_2-a_3)+\dots B_{N-1}(a_{N-1}-a_N)+a_N{B}_N\\ & =\sum _{k=1}^{N-1}{B}_k(a_k-a_{k+1})+a_N{B}_N\end{array}$$ Since {a_n} is monotone and bounded it converges; and {B_N} converges since ∑b_n converges. Hence a_NB_N converges. We estimate ∑B_k(a_k−a_k+1). Since ∑b_n converges, |∑b_n| ≤ M for some M. Using the fact that {a_n} is montone we get $\sum _1^N|B_k(a_k-a_{k+1})|\le M\sum _1^N|a_k-a_{k+1}|=M|a_1-a_{N+1}|\to M|a_1-a|$, where a_k → a. Hence ∑B_k(a_k − a_k+1) converges absolutely.

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Example 1. Suppose ∑a_n converges. Then ∑n^1/na_n converges and ∑(1+1/n)ⁿa_n converges.

Proof. It’s easy to prove that f(x) = x^1/x is decreasing for x > e, by computing the derivative of (log(x))/x. Here is the proof that (1+1/n)ⁿ increases with n.

$$\begin{array}{rl}{(1+\frac{1}{n})}^n& =1+1+\cdots +(\genfrac{}{}{0ex}{}{n}{p})\frac{1}{n^p}+\cdots +\frac{1}{n^n}\\ & =1+\cdots +(1-\frac{1}{n})\dots (1-\frac{p-1}{n})\frac{1}{p!}+\cdots +\frac{1}{n^n}\\ {(1+\frac{1}{n+1})}^{n+1}& =1+\cdots +(1-\frac{1}{n+1})\dots (1-\frac{p-1}{n+1})\frac{1}{p!}+\cdots +\frac{1}{(n+1{)}^{n+1}}.\end{array}$$ The last sum has one more (positive) term than the preceding term and the pth term of the last sum is larger than the preceding pth term since each factor is larger (subtract k/(n+1) from 1 as opposed to subtracting k/n). Hence $${(1+\frac{1}{n+1})}^{n+1}>{(1+\frac{1}{n})}^n.$$ It’s easy check using L’Hopital’s rule that (1+1/x)^x → e as x → ∞, so the sequence (1+1/n)ⁿ is monotone and bounded.

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