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? 

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

