If $\sum_{n=1}^{\infty} a_{n}$ converges, then$\sum_{n=1}^{\infty} \sin(a_{n})$ also converges If series $\sum_{n=1}^{\infty} a_{n}$ converges, prove series $\sum_{n=1}^{\infty} \sin(a_{n})$ converges  too.

Is this a series function problem or something related to? 

I tried this: 

If $a_{n} \ge 0 $ for all $n$, then $| \sin (a_{n})| \le a_{n}$, on the other hand $\sin(x)$ is continuous function in $[0,x]$ and differentiable in $(0,x)$ then exist $c \in (0,x)$ and 
$$(x-0)\cos (c)= \sin(x)-\sin(0)$$
then 
$$x\cos(c)= \sin(x)$$

but, $|\sin(x)|=|x\cos(c)|= |x| |\cos(c)| \le |x|$, thus $|\sin (x)| \le |x|$. 

We know, $|\sin (a_{n})| \le |a_{n}| $ and $\sum_{n=1}^{\infty} a_{n}$ converges, then $\sum_{n=1}^{\infty} |\sin(a_{n})|$ converges. Therefore $\sum_{n=1}^{\infty} \sin(a_{n})$ converges.

But this proof use ${a_{n}}$ positive and in the original problem I don't have this hypotesis.
This is a "Community Wiki" answer recording a comment by Daniel Fischer under the question.  The comment provides a link The set of functions which map convergent series to convergent series to a proof that the result in question is false in general, though certainly true when the $a_n$ are non-negative. My reason for writing this answer is that comments can vanish more easily than answers and can probably also be more easily overlooked.
Write $ \sum_n \sin(a_n) = \sum_n a_n + \sum_n a_n(\frac{\sin(a_n)-a_n}{a_n}) $. Since $ \sum_n a_n $ converges, $ \lim_n a_n = 0 $. Thus, for $ n $ large enough, by a Taylor series expansion, $ |\frac{\sin(a_n)-a_n}{a_n}| = O(a_n^2) $. Hence, by the triangle inequality, $ |\sum_n \sin(a_n)| \leq | \sum_n a_n | + O(\sum_n |a_n|^3) $. Finally, we use the fact that if $ a_n $ is nonnegative, then $ \sum_n |a_n|^3 $ converges whenever $ \sum_n a_n $ converges.