8
$\begingroup$

Is the series $$\sum_{n\ge 1} \frac{\sin(n^2)}{n}$$ convergent?

My thoughts so far:

1) This is an alternating series so the integration test does not work here.

2) The Weyl inequality roughly says $$\sum_{n\le N} \sin(n^2)$$ is $O(N^{1/2+\epsilon})$, so the Dirichlet test does not work directly, but one can take $$a_n=n^{-1},b_n=\sum_{k\le n} \sin(k^2)$$ and follow the idea of Dirichlet test. The problem now is that the Weyl bound does not hold for all $N$.

$\endgroup$
5
  • $\begingroup$ See also math.stackexchange.com/questions/342637/…. $\endgroup$ – Dietrich Burde Mar 27 '15 at 15:23
  • $\begingroup$ What are your thoughts so far? To get you started, notice that the sum of $(-1)^n/n$ converges while the sum of $1/n$ diverges. This sum is between them, so there is some question about how $\sin(n^2)$ behaves. If the result is true, then Dirichlet's test can probably prove it. $\endgroup$ – Ian Mar 27 '15 at 15:24
  • 2
    $\begingroup$ @Chou Thank you. I think sin(x^2) and sin(x) behave much differently. $\endgroup$ – AlgRev Mar 27 '15 at 15:49
  • $\begingroup$ Is there any known result, apart from Weyl's, about the estimates for the partial sums $\sum_{1}^{m}\sin n^{2}$? I would say you may instead make your question as seeking after the estimates for the sequence of the partial sums. $\endgroup$ – Megadeth Mar 27 '15 at 15:49
  • $\begingroup$ @Chou Good point. That is kind of what I am asking for. $\endgroup$ – AlgRev Mar 27 '15 at 15:50
5
$\begingroup$

You are on the right track. The key is to consider partial sums: $$ S_N = \sum_{n=1}^{N}\frac{\sin(n^2)}{n} $$ then find a good rational approximation of $\pi$ depending on $N$, apply Weyl bound (or Weyl differencing technique) to estimate $\sum_{n=1}^{k}e^{in^2}$ and finish through partial summation.

Details on page $11$ here (it is in Italian, hope you don't mind).

$\endgroup$
2
  • $\begingroup$ Hi, Can you give the document of this link, the link is not available. $\endgroup$ – Sansi Wu Jun 13 '18 at 3:55
  • $\begingroup$ @SansiWu: link updated. $\endgroup$ – Jack D'Aurizio Jun 13 '18 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.