# What's wrong with this proof? if $\sum a_n$ converges and $\sum b_n$ converges absolutely, then $\sum a_nb_n$ converges

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.

• Your proof (as well as your comment on possible generalizations) is indeed correct. – Sangchul Lee Apr 22 '20 at 0:29
• 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 -:) – Kavi Rama Murthy Apr 22 '20 at 0:32
• I missed the bit where you require that (at least) one of them converges absolutely. My example was irrelevant for that reason. – lulu Apr 22 '20 at 0:32
• 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. – Donlans Donlans Apr 22 '20 at 0:35
• 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. – Sangchul Lee Apr 22 '20 at 0:37

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