9
$\begingroup$

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/…. $\endgroup$ Mar 27, 2015 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, 2015 at 15:24
  • 2
    $\begingroup$ @Chou Thank you. I think sin(x^2) and sin(x) behave much differently. $\endgroup$
    – AlgRev
    Mar 27, 2015 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. $\endgroup$
    – Yes
    Mar 27, 2015 at 15:49
  • $\begingroup$ @Chou Good point. That is kind of what I am asking for. $\endgroup$
    – AlgRev
    Mar 27, 2015 at 15:50

1 Answer 1

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, 2018 at 3:55
  • $\begingroup$ @SansiWu: link updated. $\endgroup$ Jun 13, 2018 at 13:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .