Convergence of $\sum_{n=1}^{+\infty}n\tan \left( \frac{\pi}{2^{n+1}}\right )$ I am trying to find the convergence of the following series:
$$\sum_{n=1}^{+\infty}n\tan \left( \frac{\pi}{2^{n+1}}\right )$$
I am stuck trying out different tests but none of them seem to give me an answer. What do you suggest that I should try, and what are the identities or series that I can use to compare this series to so I could maybe solve it like that?
How should I approach finding the convergence of trigonometric series in general and what should I be careful of?
 A: $$
\sum_{n=1}^{+\infty}n\tan \left( \frac{\pi}{2^{n+1}}\right )
$$
The general term can be rewritten as
$$
n\tan \left( \frac{\pi}{2^{n+1}}\right ) = n\frac{\pi}{2^{n+1}}\frac{\tan \left( \frac{\pi}{2^{n+1}}\right )}{\frac{\pi}{2^{n+1}}}
$$
Use the $n$-root test
$$
\left(n\tan \left( \frac{\pi}{2^{n+1}}\right )\right)^{\frac{1}{n}} 
= \left( n\frac{\pi}{2^{n+1}}\frac{\tan \left( \frac{\pi}{2^{n+1}}\right )}{\frac{\pi}{2^{n+1}}} \right)^{\frac{1}{n}}
=  n^{\frac{1}{n}}\left(\frac{\pi}{2^{n+1}}\right)^{\frac{1}{n}}\left(\frac{\tan \left( \frac{\pi}{2^{n+1}}\right )}{\frac{\pi}{2^{n+1}}} \right)^{\frac{1}{n}} \\
$$
take the limt you get
$$
\lim_{n\to \infty} \left[n^{\frac{1}{n}}\left(\frac{\pi}{2^{n+1}}\right)^{\frac{1}{n}}\left(\frac{\tan \left( \frac{\pi}{2^{n+1}}\right )}{\frac{\pi}{2^{n+1}}} \right)^{\frac{1}{n}}\right] =  1\cdot {\frac{1}{2}} \cdot 1 = {\frac{1}{2}} < 1
$$
since the limits of the individaul factors exist, then the limit is the their product. The last one can be found by examining the limit of its logarithm.
A: Using the ratio test, you would need to find whether $$\lim_{n \to \infty} \frac{n\tan \left( \frac{\pi}{2^{n+1}}\right )}{(n-1)\tan \left( \frac{\pi}{2^n}\right )} < 1$$ is true. This simplifies to $$\lim_{n \to \infty}\frac{n}{n-1} \cdot \lim_{n \to \infty} \frac{\tan \left( \frac{\pi}{2^{n+1}}\right )}{\tan \left( \frac{\pi}{2^n}\right )}$$
The first limit is clearly $1$, and the second limit could be found by using that $\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$. This makes it $$\lim_{n \to \infty} \frac{\tan \left( \frac{\pi}{2^{n+1}}\right )}{\frac{2\tan \left( \frac{\pi}{2^{n+1}}\right )}{1-\tan^2\left( \frac{\pi}{2^{n+1}}\right )}} = \lim_{n \to \infty} \frac{1-\tan^{2}\left(\frac{\pi}{2^{n+1}}\right)}{2}$$
As $n \to \infty$, $\tan^{2}\left(\frac{\pi}{2^{n+1}}\right) \to \tan^2(0) = 0$. Then the limit for the ratio test is $$\frac{1}{2} < 1$$
Therefore, the sum converges.
In general, there won't be a nice catch-all test to determine the convergence of trig series. For most series, you should make sure that the limit of the summand is $0$. If it is, direct comparison or ratio test are my go-tos. If neither produce a clear outcome, I check the convergence using root test and integral test, and then other tests.
