2
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

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.

4
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.