# $a_n$ is convergent, $b_n$ bounded, prove $\sum a_n b_n$ converges

Suppose that $\sum_{n=0}^\infty a_n$ is a convergent series with $a_n$ > $0$ and suppose that $(b_n){_{n\in\mathbb{N}}}$ is a bounded sequence of positive numbers. Show that $\sum_{n=0}^\infty a_n b_n$ is convergent.

Since $b_n$ is bounded and $b_n$ > $0$, can we conclude that there exists M > $0$ such that $0$ < $b_n$ < M or only that $b_n$ > $0$? If we cannot conclude that $b_n$ < M then how can we answer this question?

• Yes you can conclude $0<b_n<M$. That is precisely what it means for a sequence of positive numbers to be bounded. Dec 12, 2017 at 22:51
• Okay, thanks. So then we can say that $\sum a_n b_n$ < $\sum a_n M$ = M$\sum a_n$ and therefore by the comparison test $\sum a_n b_n$ is convergent? Dec 12, 2017 at 22:54
• In fact all we need is the upper bound $M\sum a_n<+\infty$ and $\sum a_nb_n\nearrow$ which happens as $a_nb_n\ge 0$.
– zwim
Dec 12, 2017 at 22:58
• @DuncanRamage can we not assume the first inequality because it is stated that $a_n$ > $0$ and $b_n$ > $0$? Dec 12, 2017 at 23:04

Note that $$\left|\sum_{k=n+1}^ma_kb_k\right|\leq\sum_{k=n+1}^m|a_kb_k|\leq M \sum_{k=n+1}^m|a_k|\to0$$ as $m,n\to\infty$.
Since $(b_n)$ is bounded and positive, there is an $M\geq b_n >0$ for alle $n\in\mathbb{N}$.By the direct comparision test we have:
$\sum_{n=0}^\infty a_nb_n\leq \sum_{n=0}^\infty a_n\cdot M=M\underbrace{\sum_{n=0}^\infty a_n}_{<\infty}<\infty$
• @DuncanRamage It's mentioned in the post that $b_n$ are positive. Dec 12, 2017 at 22:59