Absolute convergence of the series $sin(a_n)$ implies...? Can someone please give me a counterexample for the following statement? 
If $\sum _1 ^\infty sin(a_n) $ absolutely converges , then $\sum_1^\infty a_n$ also absolutely converges.
I tried working with $a_n = \frac{ (-1)^n } {n} $ and $a_n = \frac{(-1)^n lnn}{n}$ , but without any success...
Got any idea?
Thanks a lot ! 

If $a_n=\pi$, then $\sin a_n=0$, and thus $\sum_{n=1}^\infty\sin a_n$ converges absolutely, while $\sum_{n=1}^\infty a_n$ does not.
Hence, you need to make an assumption: $a_n\in(-\pi/2,\pi/2)$.
Then use the fact that
$$
\frac{2|x|}{\pi}\le|\sin x|\le |x|,
$$
for all $x\in(-\pi/2,\pi/2)$, which implies that:
$$
\sum_{n=1}^\infty a_n\,\,\,\text{converges absolutely}\,\,\,\Longleftrightarrow
\sum_{n=1}^\infty \sin a_n\,\,\,\text{converges absolutely}.
$$
