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 $\sum_{n=0}^\infty a_nb_n$ 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,

\begin{eqnarray*} \sum_{i=m}^n a_ib_i &=& \sum_{i=0}^n a_ib_i - \sum_{i=0}^{m-1} a_ib_i\\ &=& \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_nb_n - A_{m-1}b_{m-1}\\ &=&\sum_{i=m-1}^{n-1}A_i(b_i-b_{i+1}) +A_nb_n -A_{m-1}b_{m-1}\\ |\sum_{i=m}^{n} a_ib_i|&\leq& \sum_{i=m-1}^{n-1}|A_i(b_i-b_{i+1})| + |A_nb_n| + |A_{m-1}b_{m-1}|\\ &\leq& M \sum_{i=m-1}^{n-1}(b_i-b_{i+1}) + |A_nb_n| + |A_{m-1}b_{m-1}|\\ \end{eqnarray*} 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 $m,n>N(\epsilon)$ , $|\sum_{i=m}^n a_ib_i|<\epsilon$ and $\sum a_nb_n$ converges.