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Let $x \in (0, 2\pi)$. Is the series $\sum_{n=1}^{\infty} \frac{\cos(n^2x)}{n}$ convergent? My guess is: YES and I would like to use Dirichlet test: however I have troubles proving that the partial sums $\cos(x)+\cos(4x)+...+\cos(n^2x)$ are bounded due to the lack of the simple formula for this sum. Any ideas?

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  • $\begingroup$ I'm not sure, but I don't think the partial sums are bounded. My guess is that there are some $x$ with annoying continued fraction expansion. However, I think the set of such $x$ is measure $0$. $\endgroup$ – mathworker21 Dec 10 '18 at 0:20
  • $\begingroup$ You could look at the following question first: Is there an x such that $\cos(n^2x)>\epsilon$ for all n for some epsilon >0. If that is the case (which looks likely to me even though I can‘t imediatly see such an element) you can lower bound that sum by the harmonic sum which is divergent. Therefore for that x the sum would be divergent as well $\endgroup$ – Börge Dec 10 '18 at 1:00
  • $\begingroup$ At least sometimes it is convergent: take $x=\pi$. $\endgroup$ – YiFan Dec 10 '18 at 1:07
  • $\begingroup$ And sometimes it is divergent: take $x=\frac{\pi}2$. For odd indeces $n$ the fraction vanishes, for even indeces $n$ it is $\frac 1n$. $\endgroup$ – Pavel R. Dec 10 '18 at 1:11
  • $\begingroup$ It is the Fourier series of an $L^2$ function. Carleson's theorem implies that it converges a.e. $\endgroup$ – Julián Aguirre Dec 10 '18 at 17:36

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