Prove that $\sum_{n=1}^\infty a_n b_n $ is convergent if $\sum_{n=1}^\infty a_n$ is convergent and $\sum_{n=1}^\infty (b_n -b_{n+1})$ is absolutely convergent series.
Since, $\sum_{n=1}^\infty (b_n -b_{n+1})$ converges absolutely i.e. $\sum_{n=1}^\infty \vert (b_n -b_{n+1}) \vert$ converges implies $\sum_{n=1}^\infty (b_n -b_{n+1})$ converges also.
Also $\sum_{n=1}^\infty a_n $ is also a convergent sequence.
Let, $A_n= \sum_{k=0}^n a_k $. Then for $0 \leq p \leq q $, we have
$\sum_{n=p}^q a_n b_n = \sum_{n=p}^{q-1} A_n (b_n-b_{n+1})+ A_q b_q – A_{p-1} b_p$
I think I need to use the comparison test or I need to show $\sum b_n $ is bounded.
Best Answer
The fact that $\sum_n(b_n-b_{n+1})$ converges means $b_1-b_{n+1}=\sum_{k=1}^n(b_k-b_{k+1})$ converges, hence $b_n$ converges. Similarly, $A_n:=\sum_{k=1}^na_k$ converges. Hence both sequences $A_n$ and $b_n$ are bounded by, say, $A$ and $B$ respectively, and both are Cauchy sequences.
$$\sum_{n=p}^qa_nb_n=\sum_{n=p}^qA_n(b_n-b_{n+1})+A_qb_{q+1}+b_pA_{p-1}$$ Hence \begin{align}|\sum_{n=p}^qa_nb_n|&\le |\sum_{n=p}^qA_n(b_n-b_{n+1})|+ |A_q||b_{q+1}-b_p|+|b_p||A_q-A_{p-1}|\\ &\le A\sum_{n=p}^q|b_n-b_{n+1}|+A|b_{q+1}-b_p|+B|A_q-A_{p-1}|\to0\end{align} as $p,q\to\infty$ since all terms are Cauchy. Hence the series is Cauchy and converges.