Does $\sum_{n=1}^\infty \frac{1}{n! \sin(n)}$ diverge or converge? Does the series $$ \sum_{n=1}^\infty \frac 1 {n!\sin(n)}$$ converge or diverge? Even the necessary condition of the convergence is difficult to verify.

The question is: How good can $\pi$ be approximated by rationals? From here we learn that the irrationality measure of $\pi$ is $<8$. That is: There are at most finitely many fractions $\frac nk$ with $$\tag1\left|\pi-\frac nk\right|<\frac1{k^8}.$$
For the $n$th summand in your series, $\sin n$ can only be small if $n\approx k\pi$ for some $k$.
By $(1)$, we have $|k\pi-n|>\frac1{k^7}\approx \frac{\pi^7}{n^7}\gg\frac2{n^7}$ for almost all summands. Since for small $x$ we have $|\sin x|\approx|x|$, then also $|\sin n|\gg\frac1{n^7}$. Hence we can compare the absolute series 
$$ \sum_{n=1}^\infty\left|\frac1{n!\sin n}\right|$$
with
$$ \sum_{n=1}^\infty\frac{n^7}{n!}$$
which clearly converges. Therefore your series converges absolutely.
