Convergence of $\sum_n \frac{|\sin(n^2)|}{n}$ 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.
 A: 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:
$$\sum_{n=N+1}^{\infty} \frac{|\sin(n^2)|}{n} = \sum_{k=0}^\infty \left(\sum_{n=2^k N+1}^{2^{k+1}N} \frac{|\sin(n^2)|}{n} \right)> \sum_{k=0}^\infty \frac{1}{8} = \infty$$
A: 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 $\pi$.
A: Use the limit:
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}|\sin(k^2)|=\frac{2}{\pi};\tag{*}$$
and the lemma which is implied by Stolz's theorem :


Let $\{a_n\}_{n\ge1}$ be a sequence such that $\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^na_k=a<\infty.$
Then $\lim_{n\to\infty}\frac{1}{\ln n}\sum_{k=1}^n\frac{a_k}{k}=a.$


Proof: Let $a_k=|\sin(k^2)|$, then, by $(*)$,
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^na_k=\frac2\pi,$$
by the lemma above, we konw that
$$\sum_{k=1}^{n}\frac{|\sin(k^2)|}{k}\sim\frac{2\ln n}{\pi}\quad (n\to\infty).$$
So
$$\sum_n^{\infty}\frac{|\sin(n^2)|}{n}
=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{|\sin(k^2)|}{k}=\infty.$$
