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.

Since at least one person seemed uncomfortable with the details of Hagen von Eitzen's answer, I thought it would be a nice exercise to flesh it out.
As he says, the irrationality measure says that there are only finitely many positive integers $p,q$ such that $|\pi - \frac{p}{q}| < q^{-8}$.  Since the left side is never 0 ($\pi$ is irrational), we may find a constant $c$ such that for all positive integers $p,q$, we have $|\pi - \frac{p}{q}| \ge c q^{-8}$.
To elaborate, let $\{(p_1, q_1), \dots, (p_m, q_m)\}$ be all the pairs of positive integers such that $|\pi - \frac{p_i}{q_i}| < q_i^{-8}$.  Then set $$c := \min\left\{1, q_i^8 \left|\pi - \frac{p_i}{q_i}\right| : i=1, \dots, m\right\}.$$ Since $c$ is the minimum of finitely many strictly positive numbers, we have $c>0$.
Now given a positive integer $n \ge 3$, let $k_n$ be the nearest integer to $\frac{n}{\pi}$.  Then $|n - k_n \pi| \le \frac{\pi}{2}$.  Noting that for any $|x| < \frac{\pi}{2}$, we have $|\sin x| \ge \frac{|x|}{2}$, we thus have
$$|\sin (n)| = |\sin(n-k_n\pi)| \ge \frac{|n-k_n\pi|}{2} = \frac{k_n}{2} \left| \frac{n}{k_n} - \pi \right| \ge \frac{k_n}{2} \cdot c k_n^{-8} = \frac{c}{2 k_n^7}. $$
I claim for sufficiently large $n$ we have $k_n \le \frac{n}{3}$.  Specifically, since $k_n$ is the nearest integer to $\frac{n}{\pi}$, we have $$k_n \le \frac{n}{\pi} + \frac{1}{2} = \frac{n}{3} + \frac{1}{2} - \left(\frac{1}{3}-\frac{1}{\pi}\right)n.$$  Therefore, we will have $k_n \le \frac n 3$ for any $n$ such that $\frac{1}{2} - \left(\frac{1}{3}-\frac{1}{\pi}\right)n \le 0$.  Solving this inequality for $n$, we find that we get $k_n \le \frac n 3$ if $$n \ge \frac{1}{2 \left(\frac{1}{3} - \frac{1}{\pi}\right)} \approx 33.28;$$ in particular, for all $n \ge 34$.
Hence for all $n \ge 34$ we have
$$|\sin(n)| \ge c' n^{-7}$$
where $c' = \frac{3^7 c}{2}$, and therefore
$$\left|\frac{1}{n! \sin(n)} \right| \le c' \frac{n^7}{n!}.$$
Since $\sum \frac{n^7}{n!}$ converges, we have that $\sum \frac{1}{n! \sin(n)}$ converges absolutely.
