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

Question

I am trying to find the convergence of the following series:
$$sum_{n=1}^{+infty}ntan left( frac{pi}{2^{n+1}}right )$$
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?

Varun Vejalla · Accepted Answer

Using the ratio test, you would need to find whether $$lim_{n to infty} frac{ntan left( frac{pi}{2^{n+1}}right )}{(n-1)tan left( frac{pi}{2^n}right )} < 1$$ 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 $tan(2x) = frac{2tan(x)}{1-tan^2(x)}$. This makes it $$lim_{n to infty} frac{tan left( frac{pi}{2^{n+1}}right )}{frac{2tan left( frac{pi}{2^{n+1}}right )}{1-tan^2left( frac{pi}{2^{n+1}}right )}} = lim_{n to infty} frac{1-tan^{2}left(frac{pi}{2^{n+1}}right)}{2}$$
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.

Physor · Answer

$$
sum_{n=1}^{+infty}ntan left( frac{pi}{2^{n+1}}right )
$$
The general term can be rewritten as
$$
ntan left( frac{pi}{2^{n+1}}right ) = nfrac{pi}{2^{n+1}}frac{tan left( frac{pi}{2^{n+1}}right )}{frac{pi}{2^{n+1}}}
$$
Use the $n$-root test
$$
left(ntan left( frac{pi}{2^{n+1}}right )right)^{frac{1}{n}} 
= left( nfrac{pi}{2^{n+1}}frac{tan left( frac{pi}{2^{n+1}}right )}{frac{pi}{2^{n+1}}} right)^{frac{1}{n}}
=  n^{frac{1}{n}}left(frac{pi}{2^{n+1}}right)^{frac{1}{n}}left(frac{tan left( frac{pi}{2^{n+1}}right )}{frac{pi}{2^{n+1}}} right)^{frac{1}{n}} \
$$
take the limt you get
$$
lim_{nto infty} left[n^{frac{1}{n}}left(frac{pi}{2^{n+1}}right)^{frac{1}{n}}left(frac{tan left( frac{pi}{2^{n+1}}right )}{frac{pi}{2^{n+1}}} right)^{frac{1}{n}}right] =  1cdot {frac{1}{2}} cdot 1 = {frac{1}{2}} < 1
$$
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.

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

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