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A problem in Makarov's Selected problems in real analysis asks to investigate the convergence of $\displaystyle \sum_n \frac{|\sin(n^2)|}{n}$

I'm clueless at the moment. I can't find any good property of the sequence $|\sin(n^2)|$.

$|\sin(n^2)|$ is small whenever $n\sim \sqrt{p\pi}$, and, as $p\to \infty$, the $\sqrt{p\pi}$ get closer to each other since $\sqrt{(p+1)\pi}-\sqrt{p\pi}\sim \frac 12 \sqrt{\frac{\pi}{p}}$.

Any hint is appreciated.

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  • $\begingroup$ The letter $p$ doesn't mean "prime", does it? Just in case. $\endgroup$ – ajotatxe Aug 23 '16 at 20:18
  • $\begingroup$ @ajotatxe no, just any integer $\endgroup$ – Gabriel Romon Aug 23 '16 at 20:18
  • $\begingroup$ I suspect this diverges because it's similar to $\sum_{n=1}^\infty(\frac{1}{n})$ however, that sequence unfortunately bounds our sequence of interest from above rather than below (and hence isn't useful for proving divergence). $\endgroup$ – Justin Benfield Aug 23 '16 at 20:45
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    $\begingroup$ Your series diverges with the above absolute value, and converges without it. It's a tricky one. As a hint, try considering the average value of sine (with and without absolute value) along a single period. $\endgroup$ – Shlomi A Aug 23 '16 at 20:50
  • $\begingroup$ @ShlomiA is equidistribution what you were hinting at ? $\endgroup$ – Gabriel Romon Aug 23 '16 at 21:39
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Weyl's equidistribution theorem, see e.g. Weyl-Tao, Corollary 6 implies that $n^2$ is asymptotically equidistributed mod $2\pi$. Thus as $N\rightarrow +\infty$: $$ M_N=\frac{1}{N} \sum_{n=1}^N |\sin(n^2)| \rightarrow \frac{1}{2\pi}\int_0^{2\pi} |\sin(t)|dt=\frac{2}{\pi}$$ Now, find a sequence $N_k$, $k\geq 1$, so that $10N_k\leq N_{k+1}$ and $M_{N_k}\geq \frac{1}{\pi}$. Then for every $k\geq 1$ $$ \sum_{n=N_k+1}^{N_{k+1}} \frac{|\sin n^2|}{n} \geq \frac{1}{N_{k+1}}\sum_{n=N_k+1}^{N_{k+1}} |\sin n^2| \geq M_{N_{k+1}} - \frac{N_k}{N_{k+1}} \geq \frac{1}{\pi}-\frac{1}{10}>0$$ Summing over $k$ we conclude the divergence of $\sum_{n\geq 1} |\sin(n^2)|/n$

One may note that the result (as well as the proof) goes through when replacing $n^2$ by any polynomial in $n$, $p(n)=a_d n^d+\cdots +a_0$ as long as the leading coefficient $a_d$ is rationally independent from $\pi$.

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  • $\begingroup$ Ok, thanks for the clarification, great proof! $\endgroup$ – Gabriel Romon Aug 23 '16 at 21:30
  • $\begingroup$ @H-H-Rugh, nice! Can you demonstrate that it converges without the absolute value? $\endgroup$ – Shlomi A Aug 24 '16 at 18:52
  • $\begingroup$ @ShlomiA Nope, excellent question. I have thought about it but for that the equidistribution has to be sufficiently rapid and that is not guaranteed by Weyl (as far as I know). $\endgroup$ – H. H. Rugh Aug 24 '16 at 18:57
  • $\begingroup$ @ShlomiA PS: I have posed the question on MO (I don't think it has an easy answer). $\endgroup$ – H. H. Rugh Aug 24 '16 at 19:39
  • $\begingroup$ @H-H-Rugh, how about the Dirichlet convergence test? I think that oughtta do it. $\endgroup$ – Shlomi A Aug 24 '16 at 19:42
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The series diverges. The sloppy reasoning is that $|\sin(n^2)|$ is roughly a random value between $0$ and $1$ with an average greater than or equal to $0.1$ and $\sum_n \frac{0.1}{n}$ certainly diverges.

To make this more rigorous, as @H.H.Rugh points out in his solution, as $n \to \infty$, $n^2$ is equidistributed mod $\pi$ (using Corollary 6 from here). Thus

$$|\sin(n^2)| > \frac{1}{2} \ \text{ whenever }\ n^2 \text{ mod }\pi \in \left(\frac{\pi}{6},\frac{5\pi}{6}\right)$$

which happens two thirds of the time (asymptotically). Therefore for large enough $N \in \mathbb{Z}$,

$$\sum_{n=N+1}^{2N} \frac{|\sin(n^2)|}{n}> \frac{1}{2N} \sum_{n=N+1}^{2N} |\sin(n^2)| > \frac{1}{2N} \frac{N}{2} \frac{1}{2} = \frac{1}{8}$$

since $|\sin(n^2)| > \frac{1}{2}$ for more than one half of the $n$ in $\{N+1,...,2N\}$

Therefore the series diverges since the tail of the series diverges:

$$\sum_{n=N+1}^{\infty} \frac{|\sin(n^2)|}{n} = \sum_{k=0}^\infty \left(\sum_{n=2^k N+1}^{2^{k+1}N} \frac{|\sin(n^2)|}{n} \right)> \sum_{k=0}^\infty \frac{1}{8} = \infty$$

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  • $\begingroup$ it's Corollary* and thanks for the heuristics $\endgroup$ – Gabriel Romon Aug 24 '16 at 8:44

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