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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?

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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.


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