About the convergence of $\sum_{n=1}^\infty(-1)^n\tan(\frac{\pi}{n+2})\sin(\frac{n\pi}{3} )$ Does the series $$\sum_{n=1}^\infty(-1)^n\tan\left(\frac{\pi}{n+2}\right)\sin\left(\frac{n\pi}{3}\right )$$ converge and why? I think that Leibniz' test may be helpful, but I wasn't able to find a proper way to apply it.
 A: The function $\;\tan\frac\pi{n+2}\;$ is monotone descending and positive, and also $\;\lim\limits_{n\to\infty}\tan\frac\pi{n+2}=0\;$ , and since
$$\sin\frac{n\pi}3=\begin{cases}\pm\frac{\sqrt3}2\;,&n=1,2\pmod 3\\{}\\0\,,&n=0\pmod 3\end{cases}$$  
and its series is bounded:
$$\left|\sum_{n=1}^N\sin\frac{n\pi}3\right|=\left|\frac{\sqrt3}2+\frac{\sqrt3}2+0-\frac{\sqrt3}2-\frac{\sqrt3}2-0+\frac{\sqrt3}2+\ldots\right|\le\sqrt3$$
then using Dirichlet's test the series converges. Fill in details.(for example, how do you handle the $\;(-1)^n\;$ ?)
A: You may write $(-1)^n$ as $\cos(\pi n)$, so the sequence given by:
$$ a_n = (-1)^n\sin\left(\frac{\pi}{3}n\right) = \frac{1}{2}\left[\sin\left(\frac{4\pi}{3}n\right)-\sin\left(\frac{2\pi}{3}n\right)\right]$$
has partial sums bounded by $\frac{\sqrt{3}}{2}$ in absolute value and
$$ \sum_{n\geq 1} a_n \tan\left(\frac{\pi}{n+2}\right) $$
is conditionally convergent by Dirichlet's test.
A: We can write the series as $\displaystyle\frac{\sqrt{3}}{2}\sum_{n\in S}(-1)^{r(n)}\tan\left(\frac{\pi}{n+2}\right)$, 
where $S=\{n: n\ge1, n\not\equiv0{\pmod 3}\}$ and $r(n)$ is the remainder when $n$ is divided by 3,
so the series converges by the Alternating Series Test since $\tan\left(\frac{\pi}{n+2}\right)$ is decreasing and converges to 0.
