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.

Check this: http://math.stackexchange.com/questions/1057350/limit-of-a-sequence-with-pi-and-e/1057359#1057359 — mvggz, Dec 11 '14 at 13:59

score 4 · Accepted Answer · answered Dec 11 '14 at 14:02

4

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

answered Dec 11 '14 at 14:02

user2097

1 Answers1