# Product of a Convergent Series and Bounded Sequence

Let $a_n$ be a bounded sequence and $\sum_{n=1}^\infty b_n$ be a convergent series. Then $\sum_{n=1}^\infty b_na_n$ is convergent.

I have found a counterexample to prove it false;

If we let $a_n$=$(-1)^n$ and $b_n$ = $(-1)^{n+1}$$1\over{n} Then \sum_{n=1}^\infty b_na_n diverges But if b_n$$>0$ the series converges

How can I prove this in general?

We show that if the $$b_n$$ are positive, then the series $$\sum a_kb_k$$ is absolutely convergent and hence convergent. To show absolute convergence we show that the partial sums $$\sum_{k=1}^n |a_kb_k|$$ are bounded above.
Let $$A$$ be an upper bound on the $$|a_n|$$. Then $$\sum_{k=1}^n |a_kb_k|\le A \sum_{k=1}^n |b_k|.$$ Since the $$b_k$$ are positive, and the series $$\sum b_k$$ converges to say $$B$$, we conclude that $$\sum_{k=1}^n |a_kb_k|\le AB.$$ The sequence of partial sums $$\sum_{k=1}^n |a_kb_k|$$ is non-decreasing and bounded above, so it converges.