# 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 DirichletsConvergenceTest 2013-03-22 13:19:53 2013-03-22 13:19:53 lieven (1075) lieven (1075) 5 lieven (1075) Theorem msc 40A05