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

Mathematics Asked by Dzamba on January 24, 2021

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?

2 Answers

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.

Correct answer by Varun Vejalla on January 24, 2021

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

Answered by Physor on January 24, 2021

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