<s>[INST] How find this$\lim_{n\to\infty}n^2\left(n\sin{(2e\pi\cdot n!)}-2\pi\right)=\frac{2\pi(2\pi^2-3)}{3}$ show  that 

$$\lim_{n\to\infty}n^2\left(n\sin{(2e\pi\cdot n!)}-2\pi\right)=\dfrac{2\pi(2\pi^2-3)}{3}$$

we are kown that
$$\lim_{n\to\infty}n\sin{(2\pi e\cdot n!)}=2\pi$$
because we note
$$e=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots+\dfrac{1}{n!}+\dfrac{1}{(n+1)!}+O(\dfrac{1}{(n+1)!})$$
then
$$2e\pi\cdot n! =2k\pi+\dfrac{2\pi}{n+1}+o(\dfrac{1}{n+1}))$$
for this problem we 
$$e=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots+\dfrac{1}{n!}+\dfrac{1}{(n+1)!}+\dfrac{1}{(n+2)!}+o(\dfrac{1}{(n+2)!})$$
so
$$2\pi en!=2k\pi+\dfrac{2\pi}{n+1}+\dfrac{2\pi}{(n+1)(n+2)}+o(\dfrac{1}{(n+1)(n+2)})$$
and use
$$\sin{x}=x-\dfrac{x^3}{6}+o(x^3)$$
But I can't work 
 [/INST] Clearly the fractional part of $\mathrm{e} \cdot n!$ equals to 
$$\begin{eqnarray}
  \{\mathrm{e} \cdot n!\} &=& \left\{\sum_{k=0}^{n} \frac{n!}{k!} + \sum_{k=0}^\infty \frac{n!}{(n+1+k)!}\right\} =\left\{ \sum_{k=0}^\infty \frac{n!}{(n+1+k)!}\right\} = \left\{ \sum_{k=0}^\infty \frac{1}{(n+1)_{k+1}}\right\} \\ 
  &=& \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \mathcal{o}\left(n^{-3}\right)
\end{eqnarray}
$$
Hence
$$
    n \cdot \sin\left(2 \pi \mathcal{e} n!\right) = 2 \pi - \frac{2 \pi}{n^2} \left(1 + \frac{2 \pi^2}{3}\right) + \mathcal{o}\left(n^{-3}\right)
$$
Hence 
$$
    \lim_{n \to \infty} n^2 \left( \sin\left(2 \pi \cdot \mathcal{e}\cdot n!\right) - 2 \pi \right) = - 2 \pi  \left(1 + \frac{2 \pi^2}{3}\right) \approx -47.6248875475793{\color
\gray{467109}}$$
Here is a numerical confirmation:

</s>