How to compute $\lim_{n\to \infty} n\sin(2\pi n! e)$

Question

I want to calculate

$$\lim_{n\to \infty} n\sin(2\pi n! e)$$

I have used the Stirling approximation and I think the answer is zero . But I think the limit maybe not exists.

Can some one help? Thanks.

Answer 1

$$\sin(2\pi e n!)=\sin(2\pi n!(1+1+1/2!+\ldots+1/n!+\ldots))=\sin\left(\frac{2\pi}{n+1}\right)+o(n^{-1}),$$ so the limit equals $2\pi$.