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Assume that $\sum_{n}^{\infty} a_n$ is a non-negative, convergent series. Let $f$ be a continuous function with domain $\mathbb{R}$. I have to figure out if the series

$\sum_{n}^{\infty} a_n f(\sin n)$

  1. converge
  2. diverge
  3. not enough information to decide

The only possible test that is applicable here is I think basic comparison test, somehow using the fact that sin is bounded, but I have no idea how to proceed. Could anyone help me out?

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  • $\begingroup$ What is a non-decreasing convergent series? $\endgroup$
    – Kavi Rama Murthy
    Mar 25, 2021 at 6:38
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    $\begingroup$ Converge, because $f(\sin(n))$ is bounded, that is $\sum_n^\infty a_n f(\sin(n)) \le M\sum_n^\infty a_n < \infty$ for some $M$, am I right? $\endgroup$
    – Hugo
    Mar 25, 2021 at 6:39
  • $\begingroup$ @KaviRamaMurthy non-negative means for all n, $a_n \geq 0$. $\endgroup$
    – user900404
    Mar 25, 2021 at 6:40
  • $\begingroup$ @KaviRamaMurthy, sorry, my mistake. What I mean was non-negative, not non-decreasing. $\endgroup$
    – user900404
    Mar 25, 2021 at 6:42
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    $\begingroup$ @abrakadabra_01: Note that $-1\le\sin n\le 1$ for all $n\in\mathbb{N}.$ Therefore you just need to look at $f\big{|}_{[-1, 1]}$ and by continuity (and compactness of the restricted domain) there must be some $M\ge 0$ such that $|f\big{|}_{[-1, 1]}|\le M.$ $\endgroup$
    – Bumblebee
    Mar 25, 2021 at 6:49

1 Answer 1

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Hint: $$\{f(\sin n)|n\in\mathbb N\}\subseteq f([-1,1])$$

now you can apply one of many useful properties of continuous functions on closed sets.

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