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:

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

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

1. $$\sum_{n=1}^{\infty} b_n$$ is a convergent series (given),
2. {$$a_n$$} is a monotone sequence, and
3. {$$a_n$$} 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?

• Instead of using Abel's theorem itself, use its proof, or rather summation by parts. For a partial sum, beside the two terms that are left over when using summation by part and that tend to zero, since $\sum_{1}^{m} b_n\to B$ and $a_n-a_{n+1}\to0$, you also get a sum $\sum_{r}^{s} (a_{n}-a_{n+1})\sum_{1}^n b_k$. The sum you bound using triangle inequality by $\sum_{r}^{s} |a_n-a_{n+1}||\sum_{1}^{n}b_k|$ The factor $|\sum_{1}^{n}b_k|$ is bounded by some $M$. So you get the bound $M\sum_{r}^{s}|a_{n}-a_{n+1}|$ which can be small.
– plop
Jun 16 at 14:49

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}