Prove that : $a_n>0$, Then, $\sum_{n=1}^\infty a_n$ converges iff $\sum_{n=1}^\infty \sin(a_n)$ converges. I am given the problem: Let $a_n>0$, prove $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty \sin(a_n)$ converges.
We have this problem in a homework, and I don't believe it can be true. The statement being biconditional implies that when $\sum_{n=1}^{\infty} \sin(a_n)$ converges, so too does $\sum_{n=1}^\infty a_n$, but if we take $a_n=\pi n$ or even $a_n=\pi$, then $\sum_{n=1}^\infty \sin(a_n)$ converges while $\sum_{n=1}^\infty a_n$ diverges. Is my thinking incorrect? 
 A: The statement as given is not true since the series $\sum \pi = \infty$, while $\sum \sin\pi = 0$, but one direction is true. 
First note that $0<\sin x< x$ for each $0<x<\pi$. If $\sum a_n$ converges, then $a_n\to 0$ as $n\to\infty$. Hence there is some $N\in\Bbb N$ such that for all $n\ge N$, we have $0<a_n < \pi$. Hence $0<\sin a_n < a_n$ for every $n\ge N$, so that
$$
\sum_{n\ge N}\sin a_n \le \sum_{n\ge N}a_n < \infty,
$$
so $\sum \sin a_n$ converges.
A: Here in other to make the statement true we assume 
$0<a_n<1$ otherwise the statement is heavily fails, with $a_n =n\pi$

Assume that, $0<a_n<1$ Then using the fact that, for all $x\in[-\frac{\pi}{2},\frac{\pi}{2}]$
  We have, $$\frac{2}{\pi}|x|\le |\sin x|\le |x|$$
  the claim follows

A: When $x>0$ is close to $0$ you have $x/2 < \sin x < x$ (and only the first of those two inequalities depends on $x$ being close to $0$).
And when $x<0$ is close to $0$ then you have $x < \sin x<x/2.$
So if $\sum_n a_n$ converges, then so does $\sum_n (a_n/2),$ and $\sin a_n$ is squeezed between those.
And if $a_n\to 0,$ then $\sin(a_n)$ is squeezed between $a_n$ and $0.$
The proposition is true of series in which $a_n\to0.$ However, the example you exhibit, in which $\sin(a_n) = 0$ for all $n$ while $a_n\to\infty,$ shows that if you can't say $a_n$ is close to $0$ for all but finitely many $n,$ then you can't draw that conclusion.
