How to find $\lim_{n\to\infty}n^2\left(\sin(2\pi en!)-\frac{2\pi}{n}\right)$ 
How to find$$\lim_{n\to\infty}n^2\left(\sin(2\pi en!)-\frac{2\pi}{n}\right)$$

I am very confused. I don't know how to show the limit exists, or what it is.

Recall that $e = 1 + \dfrac1{1!} + \dfrac1{2!} + \dfrac1{3!} + \cdots + \dfrac1{n!} + \dfrac1{(n+1)!} + \dfrac1{(n+2)!} + \cdots$. Hence,
$$n! e = \text{Integer} + \dfrac1{n+1} + \dfrac1{(n+1)(n+2)} + \dfrac1{(n+1)(n+2)(n+3)} + \cdots$$
Hence,
\begin{align}
\sin(2\pi n!e) & = \sin\left(\text{Integer} \times 2 \pi + \dfrac{2\pi}{n+1} + \dfrac{2 \pi}{(n+1)(n+2)} + \mathcal{O}(1/n^3) \right)\\
& = \sin\left(\dfrac{2\pi}{n+1} + \dfrac{2 \pi}{(n+1)(n+2)} + \mathcal{O}(1/n^3) \right)\\
& = \dfrac{2 \pi}{n+1} + \dfrac{2 \pi}{(n+1)(n+2)} + \mathcal{O}(1/n^3)
\end{align}
Hence,
$$\sin(2 \pi n!e) - \dfrac{2 \pi}{n} = -\dfrac{2 \pi}{n(n+1)} + \dfrac{2 \pi}{(n+1)(n+2)} + \mathcal{O}(1/n^3)$$
Hence,
$$n^2 \left(\sin(2 \pi n!e) - \dfrac{2 \pi}{n}\right) = 2 \pi \left(-\dfrac{n}{n+1} + \dfrac{n^2}{(n+1)(n+2)} + \mathcal{O}(1/n)\right)$$
Hence,
$$\lim_{n \to \infty} n^2 \left(\sin(2 \pi n!e) - \dfrac{2 \pi}{n}\right) = 2 \pi \times 0 = 0$$
