Convergence of $\sum_{n=1}^{+\infty}n\tan \left( \frac{\pi}{2^{n+1}}\right )$

Question

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?

Answer 1

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.

Answer 2

$$ \sum_{n=1}^{+\infty}n\tan \left( \frac{\pi}{2^{n+1}}\right ) $$ 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.