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.
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.