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

Question

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

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

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-infinity — Martin Sleziak, Jan 23 '15 at 11:54

score 4 · Accepted Answer · answered Dec 20 '12 at 14:37

4

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.

answered Dec 20 '12 at 14:37

Hagen von Eitzen

1 Answers1