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? 

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.
