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.