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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.

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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