Convergence of the series $\sum_{n=1}^\infty \frac{(\sin n)^n}{n}$. Please determine whether the series  $\displaystyle\sum_{n=1}^\infty \frac{(\sin n)^n}{n}$ converges. 
(Note: In Mathematica, the result tends to converge. Moreover, this is a problem mis-copied from Advanced Calculus Exam, so we don't know the difficulty of the problem [Maybe it can be solved in college mathematics]. ) 
 A: EDIT: This proof contains a mistake and I don't know if it can be saved (see the comments) - I'm leaving it up here in case it gives you an idea or if it can be corrected...

I've managed to prove the (absolute!) convergence of the series based on an idea in the comment by user @xavierm02, but my proof uses a deep (and relatively modern) result in Diophantine approximation. It appears to me that since I use the deep result only to prove that something that appears negligible is indeed negligible, and I leave some degrees of freedom in that proof, I am optimistic that there is a more elementary way to do this part.

Lemma. There is only a finite number of solutions to the inequality $\left| \sin n \right| > (\frac 1 2) ^{1/\sqrt n}$.

Proof of Lemma. I will skip some (I think simple) details in the proof. Let $\epsilon > 0$ be very small and observe the inequality $|\sin n| > 1-\epsilon$. Since $\sin$ is Lipschitz continuous and equal to $\pm 1$ exactly in the points $\pi/2 + \pi k$ for integers $k$ (and is symmetric around these points), there is some $\delta = \Theta(\epsilon)$ such that the above inequality is equivalent to: $|n - \pi/2 - \pi k| < \delta$. (That is, solutions are now in both $n$ and $k$.)
Dividing by $\pi n$ we get:
$$|\frac 1 \pi - \frac {2k+1} {2n}| < \frac \delta {\pi n}$$
This is a Diophantine approximation problem - how well can we approximate $1/\pi$ by rationals? Before we answer the question, let's plug in $\delta = \Theta(\epsilon)$ with the $\epsilon$ from the lemma. So up to some constant, we are interested in
$$|\frac 1 \pi - \frac {2k+1} {2n}| < \frac 1 n \left(1 - \left(\frac 1 2\right)^{1/\sqrt n} \right)$$
We now quote Mahler's theorem [1953]: $1/\pi$ is not a Liouville number*, meaning rational approximations are only "polynomially good": Given a rational approximation $p/q \approx 1/\pi$, the difference $p/q - 1/\pi$ is bounded (below!) by some constant power of $q$. However our Diophantine inequality requires a superpolynomial approximation, so it cannot be solved infinitely many times. Edit: This is wrong.
* Actually, Mahler's theorem is about $\pi$, but Liouville numbers are closed under the reciprocal operation.

Proposition. The series $\sum_{n=1}^\infty \frac {(\sin n)^n} n$ converges absolutely.

Proof. All except a finite amount of values of $n$ satisfy $|\sin n|^n \le \left( \frac 1 2 \right)^{\sqrt n}$, which decays at least as fast as $1/n^3$. QED.
