1
$\begingroup$

Give an example of two convergent series $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ so that the series $\sum_{n=1}^\infty a_n b_n$ diverges. Prove that it's not possible to give such an example if we also demand that one of the series $\sum_{n=1}^\infty a_n$ or $\sum_{n=1}^\infty b_n$ shall converge absolutely.

I was able to find the examples for the first part. For the secound part, it's simple if we assume one of the series converges absolutely and the other one doesen't, because then we know $$\sum_{n=1}^\infty a_n b_n \leq \sum_{n=1}^\infty a_n $$ if $\sum_{n=1}^\infty a_n$ converges absolutely and $\sum_{n=1}^\infty b_n$ converges conditionally. However i don't know how to prove it if thats not the case.

$\endgroup$
3
  • $\begingroup$ I disagree with $$\sum_{n=1}^\infty a_n b_n \leq \sum_{n=1}^\infty a_n $$ when both are absolute convergent. Counter example is $a: 1, 0, 0,\dots$ and $b: 2, 0, 0,\dots$ $\endgroup$
    – Arash
    Dec 2, 2017 at 23:35
  • $\begingroup$ We have $|a_nb_n| = |a_n||b_n|$ with $\sum |a_n|$ convergent and $b_n$ converging to 0. $\endgroup$
    – RRL
    Dec 2, 2017 at 23:42
  • $\begingroup$ Its incorrect if both are absolutely convergent, its only true if one of them is conditionally convergent. $\endgroup$
    – Pame
    Dec 3, 2017 at 9:38

1 Answer 1

0
$\begingroup$

Your inequality is wrong as you have seen in the comments. But we can do something similar. I'll write $\sum$ for $\sum_{n=1}^\infty$. Without out loss of generality assume $\sum a_n$ converges absolutely. Since $\sum b_n$ converges we have $b_n\to 0$ as $n\to\infty$ and that implies: $b_n$ is bounded. That means there is some $M>0$ such that $|b_n|<M$ for all $n$. We put everything together and get: \begin{align} \sum |a_nb_n| \leq \sum M|a_n| \end{align} So $\sum |a_nb_n|$ convergent by comparison test. Absolute convergence implies convergence and therefore $\sum a_nb_n$ is convergent.

$\endgroup$
3
  • $\begingroup$ Why is $b_n$ bounded? Unless its constantly increasing or decreasing it shouldn't be bounded? Couldn't it go to infinity for n $\in (10^{24} ,10^{40})$ or something similar? Also isn't my inequality true as long as one of them is conditionally convergent and the other is absolutely convergent? Surely if you add together positive and negative terms you will get lower values than if you just added the positive terms from the absolutely convergent series? $\endgroup$
    – Pame
    Dec 3, 2017 at 9:41
  • $\begingroup$ @Pame do you know the theorem that says if $\sum b_n$ is convergent then $b_n\to 0$ as $n\to \infty$? Secondly do you know the theorem that says convergent sequences are bounded? These theorems are important, I am sure you have seen them both in your lectures or book, since they come way before series. $\endgroup$
    – Shashi
    Dec 3, 2017 at 9:48
  • $\begingroup$ @Pame your reasoning for your inequality is wrong. Take then $b_1=20$ and $b_n=(-1)^n/n$ for $n>1$ and take $a_1=1$ and $a_n=0$ for $n>1$. Do you see the problem with your inequality? $\endgroup$
    – Shashi
    Dec 3, 2017 at 9:59

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.