Let $\{a_n\}$ be a sequence such that $\{a_nb_n\}$ is absolutely convergent for every $\{b_n\}$ that converges to 0. Prove that $\{a_n\}$ is absolutely convergent.

My thought was that since the $b_n$ converge to 0, then their terms are also getting small.

Thus, for large n, the $a_nb_n$ terms are small such that they absolutely converge.

Is this valid?

We are in Reals 2, have yet to learn about Banach spaces.


marked as duplicate by Adam Hughes, Empty, Mankind, user99914, PhoemueX Oct 1 '15 at 12:22

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  • $\begingroup$ What are you allowed to use in the proof (especially, how about bounded linear functionals and Banach spaces)? $\endgroup$ – Olorun Oct 1 '15 at 3:29
  • $\begingroup$ Hello and welcome! Your reasoning has a flaw. You must show that the $a_n$ converge - the convergence of the sequence of $a_nb_n$ is assumed. And what do you mean when you say the sequence is " absolutely convergent"? Do you perhaps mean the series? $\endgroup$ – Hans Engler Oct 1 '15 at 3:31
  • $\begingroup$ I assumed he is talking about the series $\sum a_n b_n$. This question is a duplicate, I will flag it soon. $\endgroup$ – Olorun Oct 1 '15 at 3:32
  • $\begingroup$ Olorun, this is reals 2 and we have not done any of that fun stuff yet , and also I do mean that the series converge absolutely $\endgroup$ – mjo Oct 1 '15 at 3:33
  • $\begingroup$ Since we are talking about series, then this question is indeed a duplicate as I mentioned. It has been answered in the link I provided above. $\endgroup$ – Olorun Oct 1 '15 at 3:39