# Product of a convergent series and a bounded sequence - Absolutely convergent?

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. {$$\Box$$}

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.
• I'm glad I could help. But editing to LaTeX mean little more than adding \$ at the appropriate places. Oct 30, 2018 at 21:07