TITLE: If $\sum_{n=1}^{\infty} a_{n}$ converges, then$\sum_{n=1}^{\infty} \sin(a_{n})$ also converges
QUESTION [10 upvotes]: 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.

REPLY [7 votes]: 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.