# Is $\sum_{n\ge 1} \sin(n^2)/n$ convergent?

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 $$N$$.

• Mar 27, 2015 at 15:23
• 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.
– Ian
Mar 27, 2015 at 15:24
• @Chou Thank you. I think sin(x^2) and sin(x) behave much differently. Mar 27, 2015 at 15:49
• 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. Mar 27, 2015 at 15:49
• @Chou Good point. That is kind of what I am asking for. Mar 27, 2015 at 15:50

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).