If $\sum_{n=1}^\infty a_n$ is convergent, then $\sum_{n=1}^{\infty} a_n\sin(n)$ If the series $\sum_{n=1}^\infty a_n$ is convergent (absolutely convergent or conditionally convergent), then
  $$ \sum_{n=1}^\infty a_n \sin(n) $$
  is also convergent.


Any hint? today I'm not at my best. Thanks in advance.
When the series is absouletely convergent, the proof is fairly straightforward as we have
$$
\sum_{n=1}^\infty |a_n| \geq \sum_{n=1}^\infty |a_n \sin(n)|
$$

as $ |\sin(n) | \leq 1 $ for all n. For conditionally convergent series, I'm not too sure on how it can be proven. Indeed, for a lot of conditionally convergent series, doing anything to the $a_n$ terms can cause the new series to not converge.
Hint: Using the alternating series test we know that for each monotonically decreasing sequence $a_n$ with nonnegative $a_n$ and $a_n\to 0$ the series $\sum_{n=0}^{\infty}(-1)^na_n$ converges.