Is the series $\sum_n|\sin n|^n/n$ convergent? 
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).
 A: On the OP request, here is the plot of first 10000 partial sums.

Following Terry Tao's suggestion, here is the plot of ($n$th partial sum) $+2^{\frac32}/\sqrt{\pi n}$ for $n$ up to one million:

The thick line in the beginning actually consists of high frequency oscillations - in the range up to 2000 it looks like this:

(I hope there are no rounding artifacts, I calculated everything with 100 decimal digits precision)
Next, following suggestion by j.c. in a comment below, I tried to plot the (discrete) Fourier transform of the first 10000 points; the result is this:

More precisely, height at the point with abscissa $n$ is the absolute value of the scalar product of the vector of first 10000 partial sums minus its average with the vector $\left(e^{\frac{2\pi i k}n}\right)_{1\le k\le 10000}$.
You see that $22$ and $355$, as well as $11$ ($=\frac{22}2$) and  $177.5=\frac{355}2$ are all clearly visible.
If I will have more time I will try to do the same with more data, to detect $52163$ mentioned by Terry Tao in a previous comment. I am not sure about the arbitrary phase shift that I introduced, though - I could start with $k=0$ instead of $k=1$, or any other $k$.
A: 
Semilog plot building on მამუკა ჯიბლაძე's picture, this time to $10^7$
A: Note that if $\pi$ were rational (with even numerator), then $\sin(n)$ would equal $1$ periodically, so the series would diverge.  Similarly if $\pi$ were a sufficiently strong Liouville number.  Thus, to establish convergence, one must use some quantitative measure of the irrationality of $\pi$.
It is known that the irrationality measure $\mu$ of $\pi$ is finite (indeed, the current best bound is $\mu \leq 7.60630853$).  Thus, one has a lower bound
$$ | \pi - \frac{p}{q} | \gg \frac{1}{q^{\mu+\varepsilon}}$$
for all $p,q$ and any fixed $\varepsilon>0$.  This implies that
$$ \mathrm{dist}( p/\pi, \mathbf{Z}) \gg \frac{1}{p^{\mu-1+\varepsilon}},$$
for all large $p$ (apply the previous bound with $q$ the nearest integer to $p/\pi$, multiply by $q/\pi$, and note that $q$ is comparable to $p$).  In particular, if $I \subset {\bf R}/{\bf Z}$ is an arc of length $0 < \delta < 1$, the set of $n$ for which $n/\pi \hbox{ mod } 1 \in I$ is $\gg \delta^{-1/(\mu-1+\varepsilon)}$-separated.  This implies, for any natural number $k$, that the number of $n$ in $[2^k,2^{k+1}]$ such that $|\sin(n)|$ lies in any given interval $J$ of length $2^{-k}$ (which forces $n/\pi \hbox{ mod } 1$ to lie in the union of at most two intervals of length at most $O(2^{-k/2})$) is at most $\ll 2^{k(1 - \frac{1}{2(\mu-1+\varepsilon)})}$, the key point being that this is a "power saving" over the trivial bound of $2^k$.  Noting (from Taylor expansion) that $|\sin(n)|^n \ll \exp( - j)$ if $n \in [2^k,2^{k+1}]$ and $|\sin(n)| \in [1 - \frac{j+1}{2^k}, 1-\frac{j}{2^k}]$, we conclude on summing in $j$ that
$$ \sum_{2^k \leq n < 2^{k+1}} |\sin(n)|^n \ll 2^{k(1 - \frac{1}{2(\mu-1+\varepsilon)})}$$
and hence 
$$ \sum_{2^k \leq n < 2^{k+1}} \frac{|\sin(n)|^n}{n} \ll 2^{- k\frac{1}{2(\mu-1+\varepsilon)}}.$$
The geometric series on the RHS is summable in $k$, so the series $\sum_{n=1}^\infty \frac{|\sin(n)|^n}{n}$ is convergent.  (In fact the argument also shows the stronger claim that $\sum_{n=1}^\infty \frac{|\sin(n)|^n}{n^{1-\frac{1}{2(\mu-1+\varepsilon)}}}$ is convergent for any $\varepsilon>0$.)
EDIT: the apparent numerical divergence of the series may possibly be due to the reasonably good rational approximation $\pi \approx 22/7$, which is causing $|\sin(n)|$ to be close to $1$ for $n$ that are reasonably small odd multiples of $11$.  UPDATE: I now agree with Will that it is the growth of $-2^{3/2}/\pi^{1/2} n^{1/2}$, rather than any rational approximant to $1/\pi$, which was responsible for the apparent numerical divergence at medium values of $n$, as is made clear by the updated numerics on another answer to this question.
A: 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. 
