TITLE: Product of a Convergent Series and Bounded Sequence
QUESTION [6 upvotes]: 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?

REPLY [9 votes]: 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.