Convergence of the product of a convergent and absolutely convergent series' elements. Prove that the series $\sum_{n=1}^{\infty} a_n b_n$ converges if the following conditions are met:

*

*series $\sum_{n=1}^{\infty} b_n$ converges,

*series $\sum_{n=1}^{\infty} (a_n - a_{n+1})$ absolutely converges.

I was thinking applying Abel's test, proving that if:

*

*$\sum_{n=1}^{\infty} b_n$ is a convergent series (given),

*{$a_n$} is a monotone sequence, and

*{$a_n$} is bounded.

Then the $\sum_{n=1}^{\infty} a_n b_n$ converges.
To prove the second one statement, we've to get following inequality: $a_{n+1} \leq a_n$. I've no idea how do we do that.
Third one, I think, is obtained by the fact, that the $\sum_{n=1}^{\infty} (a_n - a_{n+1})$ series converges, so $\lim_{n\to\infty} (a_n - a_{n+1}) = 0$. Is it correct?
 A: There is no way of proving that $(a_n)_{n\in\Bbb N}$ is monotonic, since it may well not be.
It's not hard to prove that you always have, when $m,n\in\Bbb N$ and $m\geqslant n$,\begin{multline}\sum_{k=n}^na_kb_k=(a_n-a_{n+1})b_n+(a_{n+1}-a_{n+2})(b_n+b_{n+1})+\cdots+\\+(a_m-a_{m+1})\left(\sum_{k=n}^mb_k\right)+a_{m+1}\left(\sum_{k=n}^mb_k\right).\end{multline}Note that the sequence $(a_n)_{n\in\Bbb N}$ converges. In fact\begin{align}\sum_{n=1}^\infty|a_n-a_{n+1}|\text{ converges}&\implies\sum_{n=1}^\infty(a_n-a_{n+1})\text{ converges}\\&\iff(a_n)_{n\in\Bbb N}\text{ converges.}\end{align}Since it converges, it is a bounded sequence and therefore the sequence $\bigl(|a_n|\bigr)_{n\in\Bbb N}$ is bounded too.
Now, let $\varepsilon>0$. I will prove that there is a natural number $N$ such that if $m\geqslant n\geqslant N$, then$$\left|\sum_{k=n}^ma_kb_k\right|<\varepsilon,\tag1$$in order to apply the Cauchy's convergence test. Take $\varepsilon'\in(0,\varepsilon)$ and take $N\in\Bbb N$ such that$$m\geqslant n\geqslant N\implies\left|\sum_{k=n}^mb_k\right|<\frac{\varepsilon'}{\displaystyle\sum_{k=1}^\infty|a_k-a_{k+1}|+\sup_{n\in\Bbb N}|a_n|}.$$Actually, this doesn't make sense if every $a_n$ is $0$, but then the statement is trivially true. Now we have, thanks to the first equality of this proof, that, if $m\geqslant n\geqslant N$,\begin{align}\left|\sum_{k=n}^na_kb_k\right|&\leqslant\frac{\varepsilon'}{\displaystyle\sum_{k=1}^\infty|a_k-a_{k+1}|+\sup_{n\in\Bbb N}|a_n|}\left(\sum_{k=n}^m|a_k-a_{k+1}|+\sup_{n\in\Bbb N}|a_n|\right)\\&\leqslant\varepsilon'\\&<\varepsilon.\end{align}
