4
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Your proof (as well as your comment on possible generalizations) is indeed correct. $\endgroup$ – Sangchul Lee Apr 22 '20 at 0:29
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.