Proof:
Since $\sum a_n$ converges, then $a_n \to 0 $ so there exists $M > 0$ such that $a_n < M$ for all $n$.
Also, there exists $N$ such that for all $m,n \geq N$ we get $|\sum_{k=n}^m |b_k|| < \epsilon/M$
thus $$|a_nb_n + \dots + a_mb_m| \leq |a_nb_n| + \dots + |a_m||b_m| \leq M(|b_n| +\dots +|b_m|) < M \frac{\epsilon}{M} = \epsilon$$ So by the cauchy principle, convergence follows.
I think that my proof is wrong, but I don' see the error, I think it's wrong because I only used the convergence of $\sum a_n$ to deduce that $a_n$ goes to $0$. so that hypothesis is unnecessary, according to my proof, it's enough for $a_n$ to be bounded. so $a_n$ doesn't even need to be convergent. Can someone point out the mistake please? I've tried to find a counter example but I haven't found any yet.
Best Answer
The only flaw is in the 1st sentence,where $a_n<M$ should be $|a_n|<M.$