Does the sequence $\frac{1}{e^k \sin{k}}$ converge?
If $\sin{k}$ acts as a random variable (taking on values in $(-1, 1)$), then it seems like we should be able to prove that the sequence converges with high probability. I wonder if it can be decided absolutely.
On
NOT A PROOF. But here are my thoughts. I don't think it will converge. As k continues to get larger, it will hit closer and closer multiples of pi. For example, k=3, 31, 314, 3141, 31415, ... etc. As these values of k occur, sin(k) will get closer and closer to zero. Since sin(k) is in the denominator, the effect will cause the $k^{th}$ sequence element to get bumpy, which happens when one divides by zero (or numbers close to zero). Yes, there is the $e^{k}$ in the denominator that makes the sequence appear to converge quickly, but again those approximate multiples of pi become more accurate as $k \rightarrow \infty$.
This picture shows what happens at $x=\pi, 2\pi,...$ , and what will happen as the values of k get closer to a multiple of $\pi$.
It converges to $0$, in fact $\frac{1}{k^7\sin{k}}$ already converges to $0$, see Theorem 2 here. This theorem gives a nice characterization of the irrationality measure of $\pi$ as the borderline number $\mu$ such that $\frac{1}{k^{u-1}\sin{k}}$ converges to $0$ for $u>\mu$, and diverges for $u<\mu$. So $\frac{1}{k^7\sin{k}}$ converges because $\mu$ is known to be less than $8$, and $\frac{1}{k^{1/2}\sin{k}}$ diverges because it has to be at least $2$. Whether $\frac{1}{k^2\sin{k}}$ converges is an open question, but the conventional wisdom is that $\mu=2$.
Interestingly enough, the sequence $\frac{1}{k^{u-1}\xi_k}$, where $\xi_k$ are independent random variables uniformly distributed on $[-1,1]$ converges to $0$ almost surely if and only if $\sum\frac{1}{k^{u-1}}<\infty$. So $\sin{k}$ being a "random variable" uniformly distributed on $[-1,1]$ is in fact equivalent to $\mu=2$.
Convergence of similar sequences $\frac{\tan k}{k^{u-1}}$ is analysed here, also implicitly using the irrationality measure of $\pi$.