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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.


marked as duplicate by Simon S, Thomas Andrews, Henry, Aditya Hase, Davide Giraudo real-analysis Dec 11 '14 at 14:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


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


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