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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?

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  • $\begingroup$ Yes you can conclude $0<b_n<M$. That is precisely what it means for a sequence of positive numbers to be bounded. $\endgroup$ Dec 12, 2017 at 22:51
  • $\begingroup$ 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? $\endgroup$
    – user396611
    Dec 12, 2017 at 22:54
  • $\begingroup$ In fact all we need is the upper bound $M\sum a_n<+\infty$ and $\sum a_nb_n\nearrow$ which happens as $a_nb_n\ge 0$. $\endgroup$
    – zwim
    Dec 12, 2017 at 22:58
  • $\begingroup$ @DuncanRamage can we not assume the first inequality because it is stated that $a_n$ > $0$ and $b_n$ > $0$? $\endgroup$
    – user396611
    Dec 12, 2017 at 23:04

2 Answers 2

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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 $m,n\to\infty$.

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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:

$\sum_{n=0}^\infty a_nb_n\leq \sum_{n=0}^\infty a_n\cdot M=M\underbrace{\sum_{n=0}^\infty a_n}_{<\infty}<\infty$

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    $\begingroup$ @DuncanRamage It's mentioned in the post that $b_n$ are positive. $\endgroup$
    – Math Lover
    Dec 12, 2017 at 22:59
  • $\begingroup$ I actualy do not see, why this would "prove" that non-convergent series converge. What has to be repaired? $\endgroup$
    – Cornman
    Dec 12, 2017 at 23:04
  • $\begingroup$ @Cornman I misread the question and didn't see that the an are positive. Sorry! $\endgroup$ Dec 12, 2017 at 23:12

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