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

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

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    See also http://math.stackexchange.com/questions/76097/what-is-the-limit-of-n-sin-2-pi-cdot-e-cdot-n-as-n-goes-to-infinity2015-01-23

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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.