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.

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    $\begingroup$ Your proof (as well as your comment on possible generalizations) is indeed correct. $\endgroup$ – Sangchul Lee Apr 22 '20 at 0:29
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    $\begingroup$ If $(a_n)$ is bounded and $\sum b_n$ converges absolutely then so does $\sum a_nb_n$. You can be absolutely sure about this -:) $\endgroup$ – Kavi Rama Murthy Apr 22 '20 at 0:32
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    $\begingroup$ I missed the bit where you require that (at least) one of them converges absolutely. My example was irrelevant for that reason. $\endgroup$ – lulu Apr 22 '20 at 0:32
  • $\begingroup$ So I guess my professor wrote $\sum a_n$ convergent to make it easier for us or something, I'm insecure about writing proof. Thanks for the answer. $\endgroup$ – Donlans Donlans Apr 22 '20 at 0:35
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    $\begingroup$ For a quicker proof (assuming that the comparison test has been established), you may argue that the right-hand side of $$\sum_n |a_n b_n|\leq \Bigl(\sup_n |a_n|\Bigr)\sum_n |b_n|$$ is finite, and so, $\sum a_n b_n$ converges absolutely. This is an instance of the far-reaching generalization called the Hölder's inequality. $\endgroup$ – Sangchul Lee Apr 22 '20 at 0:37

The only flaw is in the 1st sentence,where $a_n<M$ should be $|a_n|<M.$


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