# Given that $\sum_{n}^{\infty} a_n$ converge, does $\sum_{n}^{\infty} a_n f(\sin n)$ converge?

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?

• What is a non-decreasing convergent series? – Kavi Rama Murthy Mar 25 at 6:38
• 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? – Hugo Mar 25 at 6:39
• @KaviRamaMurthy non-negative means for all n, $a_n \geq 0$. – user900404 Mar 25 at 6:40
• @KaviRamaMurthy, sorry, my mistake. What I mean was non-negative, not non-decreasing. – user900404 Mar 25 at 6:42
• @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 $|f\big{|}_{[-1, 1]}|\le M.$ – Bumblebee Mar 25 at 6:49

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