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I am very confused and curious about the convergence of $\sum^{\infty}_{n=1} \tan(nx)/n^2$ when $x$ is any real number. Of course if $x=\pi/2$, the series blow up. However, what about other cases? I cannot find a way to approach...Could anyone please help me?

Hendrix
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Keith
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    My suspicion is that it won't converge except for $x = 0$. This is because $\tan(nx) = \frac{\sin(nx)}{\cos(nx)}$ and for any non-zero values of $x$, there will always be certain values of $n$ where $nx$ becomes arbitrarily close to $m\pi + \frac{\pi}{2}$ for integer $m$ and, thus, $\cos(nx)$ would be very close to $0$ so, even for $n$ very large, the value of $\frac{\tan(nx)}{n^2}$ would become substantial enough that the sum would not converge. However, even if this is true, I'm not quite sure how to show this rigorously. – John Omielan Aug 25 '19 at 03:25
  • Yes I have almost the same suspicion, same that cannot prove rigorously – Keith Aug 25 '19 at 03:26
  • @JohnOmielan No. There are other "exceptional" values of $x$, such as $\frac{\pi}{3}$. – mathworker21 Aug 25 '19 at 03:27
  • @Keith This is very likely the main issue to prove, apart from other exceptional values as mathworker21 just stated. I suggest you edit your question text to mention this issue explicitly. – John Omielan Aug 25 '19 at 03:28
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    The convergence of $\sum_n \tan(nx)n^{-a},\sum_n |\tan(nx)|n^{-a}$ is about the irrationality measure of $x/\pi$ – reuns Aug 25 '19 at 03:34

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