# How to compute $\lim_{n\to \infty} n\sin(2\pi n! e)$ [duplicate]

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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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 11 '14 at 14:28

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