TITLE: Is $\sum\limits_{n=1}^\infty \sin{\frac{(-1)^{n+1}}{n}}$ convergent?
QUESTION [5 upvotes]: $$
\mbox{Is}\quad
\sum_{n=1}^\infty \sin\left(\left[-1\right]^{n + 1} \over n\right)
\quad\mbox{ convergent ?.}
$$ 
$$
\mbox{I know that }\quad\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\quad
\mbox{is convergent.}
$$  
$\frac{\sin x}x $  goes to 1 as x goes to $0$. So I feel that the series should be convergent, but  I can't prove it rigorously.

REPLY [9 votes]: Yes. Sine is an odd function, so your summation is just:
$$\sum_n (-1)^n\sin \left({1\over n}\right)$$
which converges by the alternating series test.

REPLY [4 votes]: Hint: 
$$\sum_{n=1}^{\infty}\sin\left(\frac{(-1)^{n+1}}{n}\right)=\sum_{n=1}^{\infty}(-1)^{n+1}\sin\frac{1}{n}$$

REPLY [3 votes]: Hint
Use the Taylor's formula:
$$\sin x=_0 x+\mathcal O(x^3)$$
and you find that the given series is the sum of two convergent series: one by the Leibniz theorem and the second by comparison with a convergent Riemann series.

REPLY [2 votes]: Since 
$$\sum_{n=1}^\infty\sin\frac{(-1)^{n+1}}{n}=\sum_{n=1}^\infty(-1)^{n+1}\sin\frac{1}{n},$$
this is a alternate series. From the monotonicity (decreasing) of $\sin\frac{1}{n}$ and the limit
$$\lim_{n\to\infty}\sin\frac{1}{n}=0,$$
we see that the series is convergent.