TITLE: How to find $\lim_{n\to\infty}n\cdot\sin(2\pi\ e\ n!)$?
QUESTION [8 upvotes]: What will be the value of this limit?
$$\lim_{n\to\infty}n\cdot\sin(2\pi\ e\ n!)$$
Any help on how to proceed?

REPLY [11 votes]: Consider that, as $n$ increases, the distance of $e n!$ from the closest integer drops to zero, since:
$$ n!e=n!\cdot\sum_{k=0}^{+\infty}\frac{1}{k!}=A+\frac{1}{(n+1)}+\frac{1}{(n+1)(n+2)}+\ldots $$
where $A\in\mathbb{N}$. This gives:
$$ \{n! e\} = \frac{1}{n+1}+O\left(\frac{1}{n^2}\right)$$
and:
$$n\sin(2\pi n! e) = n\sin\left(2\pi\left(\frac{1}{n+1}+O\left(\frac{1}{n^2}\right)\right)\right),$$
so the limit is $2\pi$.