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What is the limit of $n \sin (2 \pi \cdot e \cdot n!)$ as $n$ goes to infinity?

In order to solve the following limit $\lim_{n\to\infty} n\sin2\pi n!e$ . This question is very likely to have been asked.

I remember this question and the answer is like $2\pi$ or something .

I also do remember approximating $n!e$ but somehow i don't remember and can't figure out right now .

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    Nice problem, but I'd also like to see a different approach from the classical one.2012-08-01

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Look at the usual series for $e$. Then $n!e$ is an integer plus $\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots.\tag{$1$}$ The sum $(1)$ is bigger than $\frac{1}{n+1}$. It is less than $\frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots,$ a geometric series with sum $\frac{1}{n}$.

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$ \begin{align} n!e &=n!\sum_{k=0}^\infty\frac1{k!}\\ &=\sum_{k=0}^n\frac{n!}{k!}+\sum_{k=n+1}^\infty\frac1{k!/n!}\\ &\equiv\frac1{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\dots\pmod{\mathbb{Z}} \end{align} $ where the last sum is greater than $\frac1{n+1}$ yet less than $\frac1n=\frac1{n+1}+\frac1{(n+1)^2}+\frac1{(n+1)^3}+\dots$

Thus, we have the bounds $ \frac1{n+1} Therefore, $ n\sin\left(\frac{2\pi}{n+1}\right) and by the Squeeze Theorem, we get $ \lim_{n\to\infty}n\sin(2\pi n!e)=2\pi $

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    @robjohn: very detailed! (+1)2012-08-01
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Remember that $\mathrm e=\sum\limits_{k=0}^n1/k!+R_n$ with $R_n=\sum\limits_{k=n+1}^{+\infty}1/k!$. Hence $n!\,\mathrm e$ equals an integer plus $n!R_n$. Now $1/(n+1)!\lt R_n\lt1/(n\,n!)$, hence $n!R_n=1/n+o(1/n)$, which is all the precision you need.