Dirichlet’s convergence test

Theorem. Let $\{a_{n}\}$ and $\{b_{n}\}$ be sequences of real numbers such that $\{\sum_{i=0}^{n}a_{i}\}$ is bounded and $\{b_{n}\}$ decreases with $0$ as limit. Then $\sum_{n=0}^{\infty}a_{n}b_{n}$ converges.

Proof. Let $A_{n}:=\sum_{i=0}^{n}a_{n}$ and let $M$ be an upper bound for $\{|A_{n}|\}$ . By Abel’s lemma,

$\displaystyle\sum_{i=m}^{n}a_{i}b_{i}$	$\displaystyle=$	$\displaystyle\sum_{i=0}^{n}a_{i}b_{i}-\sum_{i=0}^{m-1}a_{i}b_{i}$
	$\displaystyle=$	$\displaystyle\sum_{i=0}^{n-1}A_{i}(b_{i}-b_{i+1})-\sum_{i=0}^{m-2}A_{i}(b_{i}-% b_{i+1})+A_{n}b_{n}-A_{m-1}b_{m-1}$
	$\displaystyle=$	$\displaystyle\sum_{i=m-1}^{n-1}A_{i}(b_{i}-b_{i+1})+A_{n}b_{n}-A_{m-1}b_{m-1}$
$\displaystyle\|\sum_{i=m}^{n}a_{i}b_{i}\|$	$\displaystyle\leq$	$\displaystyle\sum_{i=m-1}^{n-1}\|A_{i}(b_{i}-b_{i+1})\|+\|A_{n}b_{n}\|+\|A_{m-1}b_{% m-1}\|$
	$\displaystyle\leq$	$\displaystyle M\sum_{i=m-1}^{n-1}(b_{i}-b_{i+1})+\|A_{n}b_{n}\|+\|A_{m-1}b_{m-1}\|$

Since $\{b_{n}\}$ converges to $0$ , there is an $N(\epsilon)$ such that both $\sum_{i=m-1}^{n-1}(b_{i}-b_{i+1})<\frac{\epsilon}{3M}$ and $b_{i}<\frac{\epsilon}{3M}$ for $m,n>N(\epsilon)$ . Then, for $m,n>N(\epsilon)$ , $|\sum_{i=m}^{n}a_{i}b_{i}|<\epsilon$ and $\sum a_{n}b_{n}$ converges.

Title	Dirichlet’s convergence test
Canonical name	DirichletsConvergenceTest
Date of creation	2013-03-22 13:19:53
Last modified on	2013-03-22 13:19:53
Owner	lieven (1075)
Last modified by	lieven (1075)
Numerical id	5
Author	lieven (1075)
Entry type	Theorem
Classification	msc 40A05