convergente of the sum of sines of the terms of the alternating harmonic series I want to know about the convergence or divergence of the following series:
$$\sum \sin (a_n) $$
where 
$$a_n=\frac{(-1)^n}{n}$$
The tests that I tried were inconclusive. Is it possible to know?
Or, more generally, does this series converge for all conditionally convergent series given by the sequence $a_n$? I know this result holds for any absolutely convergent series (here),  but I want to know if it works for any convergent series or if there's an explicit counterexample taking conditionally convergent series.
EDIT: as pointed out at the answer section, my example converges, but the general case is still unanswered. 
 A: For the first part 

Does $\sum\limits_{n=1}^\infty \sin\frac{(-1)^n}{n}$ converges?

The answer is YES by 
Alternating series test. 
For the second part

Does $\sum\limits_{n=1}^\infty \sin a_n$ converges for all conditional converging series $\sum\limits_{n=1}^\infty a_n$ ?

The answer is NO. Consider the series
$$a_n = \frac{\epsilon_n}{\lceil n/3 \rceil ^{1/3}}
\quad\text{ where }\quad
\epsilon_n = \begin{cases} 
+1,  & n \not\equiv 0 \pmod 3\\
-2, & n \equiv 0 \pmod 3
\end{cases}$$
If we group the terms in units of three, one find
$$a_{3k-2} + a_{3k-1} + a_{3k} = 0$$
and the series $\sum\limits_{n=1}^\infty a_n$ converges conditionally.
However,
$$\sin a_{3k-2} + \sin a_{3k-1} + \sin a_{3k} = 2\sin\frac{1}{\sqrt[3]{k}} - \sin\frac{2}{\sqrt[3]{k}} = \frac{1}{k} + O\left(\frac{1}{k^{5/3}}\right)$$
This implies the partial sums $\sum\limits_{n=1}^N a_{n}$ behaves roughly as $\sum\limits_{k=1}^{\lfloor N/3 \rfloor} \frac{1}{k} \sim \log\frac{N}{3}$ for large $N$
and hence the corresponding series diverges.
A: $\sin(a_n)=(-1)^n\sin(\frac{1}{n})$ so you are dealing with an alternating series. Then the fact that $\sin(\frac{1}{n})$ is monotonically decreasing and converges to $0$ justifies the conclusion that the series is converging. The so-called Leibniz-test is applied.
A: For $x$ close to $0$, we have $\frac{|x|}{2}\leq |\sin{x}|\leq |x|$. Does that give you any ideas?
Edit: can be done way simpler, just noticed. $\sin(-x)=-\sin(x)$, and the sequence $\frac{1}{n}$ has the limit $0$. Now, if I were to mention a guy called Leibniz, we are almost done.
