# How to show that $\lim_{n\to \infty}n\sin(2\pi en!)=2\pi$?

Please give me some hint to proceed. I'm clueless:

Show that, $\lim_{n\to \infty}n\sin(2\pi en!)=2\pi$

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Note that $e=\sum_{k=0}^\infty\frac1{k!}$, hence $en!=\sum_{k=0}^n\frac{n!}{k!}+\frac1{n+1}+\ldots$ is quite close to an integer.