How to find the following limit $$\lim_{n \rightarrow \infty} n \sin(2 \pi e n!)$$
Its answer is $2 \pi$. I am not seeing any clue to find it out.
Thank you for your help. If it discussed earlier please mention link.
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1http://math.stackexchange.com/questions/76097/what-is-the-limit-of-n-sin-2-pi-cdot-e-cdot-n-as-n-goes-to-infinity – lab bhattacharjee Nov 10 '13 at 16:32
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Thank you for this link. Prof. Scott's answer is much shorter. – Supriyo Nov 10 '13 at 17:09
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HINT: Recall that $e^x=\sum_{k\ge 0}\frac{x^k}{k!}$, so $e=\sum_{k\ge 0}\frac1{k!}$, and
$$en!=\sum_{k\ge 0}\frac{n!}{k!}=\sum_{k=0}^n\frac{n!}{k!}+\sum_{k\ge n+1}\frac{n!}{k!}\;.\tag{1}$$
The first term on the righthand side of $(1)$ is an integer, and the second is
$$\begin{align*} \sum_{k\ge n+1}\frac{n!}{k!}&=\frac1{n+1}+\frac1{(n+1)(n+2)}+\ldots\\\\ &\le\frac1{n+1}+\left(\frac1{n+1}\right)^2+\ldots\\\\ &=\sum_{k\ge 1}\left(\frac1{n+1}\right)^k\\\\ &=\frac{\frac1{n+1}}{1-\frac1{n+1}}\\\\ &=\frac1n\;. \end{align*}$$
Brian M. Scott
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Thank you Sir. The next part will be $\lim n \sin(2 \pi e n!) = \lim n \sin (\frac{2\pi}{n}) = \lim 2\pi \frac{\sin(\frac{2\pi}{n})}{\frac{2\pi}{n}}=2\pi$. – Supriyo Nov 10 '13 at 17:06
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@Samprity: Almost: that righthand summation is between $\frac1{n+1}$ and $\frac1n$, so $\sin 2\pi en!$ is trapped between $\sin\frac{2\pi}{n+1}$ and $\sin\frac{2\pi}n$. Now use the squeeze theorem. – Brian M. Scott Nov 10 '13 at 17:14