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.

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