TITLE: Is the series $\sum_n|\sin n|^n/n$ convergent?
QUESTION [102 upvotes]: Problem. Is the series $$\sum_{n=1}^\infty\frac{|\sin(n)|^n}n$$convergent?

(The problem was posed on 22.06.2017 by Ph D students of H.Steinhaus Center of Wroclaw Polytechnica. The promised prize for solution is  "butelka miodu pitnego", see page 37 of Volume 1 of the Lviv Scottish Book.
To get the prize, write to the e-mail: hsc@pwr.edu.pl).

REPLY [10 votes]: Let $D_N$ be the discrepancy:
$$
D_N=\sup \left| \frac{ A(J:P)}{N} - |J|\right|
$$
where $P=\{k/\pi \ \mathrm{mod} \ 1\}_{k=1,2,\ldots, n}$, $J$ is an interval in $[0,1]$. 
If the irrationality measure $\mu$ of $\pi$ is finite, then we have
$$
D_N\ll N^{-\frac1{\mu-1} + \epsilon}.
$$
From this result and Terry Tao's answer, the number of $n\in [2^k, 2^{k+1}]$ for which $|\sin n |$ falls in an interval of length $2^{-k}$, is 
$$
\ll 2^{\frac k2}  + 2^{k\left(1-\frac1{\mu-1} + \epsilon\right)}
$$
Thus, if $r>\max\left\{\frac12, 1-\frac1{\mu-1} \right\}$, then the series 
$$
\sum_{n=1}^{\infty} \frac{|\sin n|^n}{n^r}
$$
is convergent. 
It is conjectured that $\mu=2$. If we prove that $2\leq \mu <3$, then we can also prove that
$$
\sum_{n=1}^{\infty} \frac{|\sin n|^n}{\sqrt n}
$$
diverges. I am not aware of any unconditional proof of the divergence of this series.