Proving the series $\sum_{n=0}^{\infty}\sin(e\pi n!)$ converges I didn't know how to solve this problem. 
How to prove $$\sum_{n=0}^{\infty}\sin(e\pi n!)$$ converges?
 A: Edit. I eliminated the use of asymptotic notation (for those who are not acquainted with  this) and tried to be very specific. Hope this dusts out all the suspicions.

Using the fact that $e$ admits the expansion $ e = \sum_{k=0}^{\infty} \frac{1}{k!}$, for $n \geq 2$ let us write
$$ n!e = \underbrace{\sum_{k=0}^{n-2} \frac{n!}{k!}}_{=: b_n} + n + 1 + \underbrace{\sum_{k=1}^{\infty} \frac{1}{(n+1)\cdots (n+k)}}_{=: c_n}. $$
Now notice that


*

*In each term of $b_n$, we have $\frac{n!}{k!} = n(n-1)\cdots(k+1)$. Since either $n$ or $n-1$ is always even, it follows that $\frac{n!}{k!}$ is always an even integer. So the same is true for the entire sum $b_n$.

*The sequence $(c_n)$ is monotone decreasing and tends to $0$. Indeed,
$$ c_{n+1}
= \sum_{k=1}^{\infty} \frac{1}{(n+2)\cdots(n+k+1)}
\leq \sum_{k=1}^{\infty} \frac{1}{(n+1)\cdots(n+k)}
= c_n $$
and
$$ 0 \leq c_n \leq \sum_{j=1}^{\infty} \frac{1}{(n+1)^j} = \frac{1}{n}. $$
Combining all the efforts, we find that
$$ \bbox[border:2px dashed green,10px]{ \sin(e\pi n!) = (-1)^{n+1}\sin(\pi c_n)} $$
But since we know that $0 \leq \pi c_n \leq \frac{\pi}{2}$ for $n \geq 2$ and the sine function is increasing on $[0, \frac{\pi}{2}]$, the sequence $(\sin(\pi c_n))_{n\geq 2}$ is also monotone decreasing and tends to $0$. Therefore by the alternating series test, the series in question converges.
