Fractional part of $n!e$ 
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What is the limit of $n \sin (2 \pi \cdot e \cdot n!)$ as $n$ goes to infinity? 

In order to solve the following limit $$\lim_{n\to\infty} n\sin2\pi n!e$$ . 
This question is very likely to have been asked. 
I remember this question and the answer is like $2\pi$ or something .
I also do remember approximating $n!e$ but somehow i don't remember and can't figure out right now .  
 A: Look at the usual series for $e$. Then $n!e$ is an integer plus
$$\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots.\tag{$1$}$$
The sum $(1)$ is bigger than $\frac{1}{n+1}$. 
It is less than
$$\frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots,$$
a geometric series with sum $\frac{1}{n}$.
A: $$
\begin{align}
n!e
&=n!\sum_{k=0}^\infty\frac1{k!}\\
&=\sum_{k=0}^n\frac{n!}{k!}+\sum_{k=n+1}^\infty\frac1{k!/n!}\\
&\equiv\frac1{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\dots\pmod{\mathbb{Z}}
\end{align}
$$
where the last sum is greater than $\frac1{n+1}$ yet less than $\frac1n=\frac1{n+1}+\frac1{(n+1)^2}+\frac1{(n+1)^3}+\dots$
Thus, we have the bounds
$$
\frac1{n+1}<n!e-\lfloor n!e\rfloor<\frac1n
$$
Therefore,
$$
n\sin\left(\frac{2\pi}{n+1}\right)<n\sin(2\pi n!e)<n\sin\left(\frac{2\pi}{n}\right)
$$
and by the Squeeze Theorem, we get
$$
\lim_{n\to\infty}n\sin(2\pi n!e)=2\pi
$$
A: Remember that $\mathrm e=\sum\limits_{k=0}^n1/k!+R_n$ with $R_n=\sum\limits_{k=n+1}^{+\infty}1/k!$. Hence $n!\,\mathrm e$ equals an integer plus $n!R_n$. Now $1/(n+1)!\lt R_n\lt1/(n\,n!)$, hence $n!R_n=1/n+o(1/n)$, which is all the precision you need.
