Abbott's Understanding Analysis - Exercise 2.7.6.(a): Show that if $\sum_{n=1}^\infty a_n$ converges absolutely and $b_n$ is bounded, then $\sum_{n=1}^\infty a_nb_n$ converges.

Does $\sum_{n=1}^\infty a_nb_n$ converge absolutely? I've written a proof that shows that it does, but no place mentions it. Is the proof correct?

The proof:

Let $M\in\mathbb{R}$ such that $\forall n \in\mathbb{R}:M>b_n$. ($b_n$ is bounded). Let $\sum_{n=1}^\infty |a_n| = A$.

$\sum_{n=1}^\infty |a_nb_n| \le \sum_{n=1}^\infty |Ma_n| = M\sum_{n=1}^\infty |a_n| = MA$.

$\sum_{n=1}^\infty |a_nb_n|$ is bounded and monotonically ascending, therefore it is convergent.

$\sum_{n=1}^\infty |a_nb_n|$ is convergent, therefore $\sum_{n=1}^\infty a_nb_n$ is absolutely convergent. {$\Box$}

Thanks in advance.


1 Answer 1


It is almost correct. There is, however, a small flaw. When you write that $M$ is such that $(\forall n\in\mathbb{N}):b_n\leqslant M$, you should actually have written that $M$ is such that $(\forall n\in\mathbb{N}):\lvert b_n\rvert\leqslant M$, because that's what you use later.

  • $\begingroup$ True. Thank you for your answer and especially for editing the entire post to LaTeX! $\endgroup$
    – JohnShn
    Oct 30, 2018 at 21:03
  • $\begingroup$ I'm glad I could help. But editing to LaTeX mean little more than adding $ at the appropriate places. $\endgroup$ Oct 30, 2018 at 21:07

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