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?

  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.