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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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marked as duplicate by Davide Giraudo, Ross Millikan, Did, Rahul, J.D. Jul 27 '12 at 21:45

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.

Is it $\sin(2 \pi n! e)$? I would think so, but parentheses would be appreciated. – Ross Millikan Jul 27 '12 at 21:30
@Belgi: Huh? $ $ – Did Jul 27 '12 at 21:30
@Begi : I don't think that way it can be done or may be too hard . – Theorem Jul 27 '12 at 21:32
My mistake, my comment was about " but somehow i don't remember", I thought the PO doesn't remember the approximation... – Belgi Jul 27 '12 at 21:33
Nice problem, but I'd also like to see a different approach from the classical one. – user 1618033 Aug 1 '12 at 13:21

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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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.

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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 $$ \frac1{n+1}<n!e-\lfloor n!e\rfloor<\frac1n $$ Therefore, $$ n\sin\left(\frac{2\pi}{n+1}\right)<n\sin(2\pi n!e)<n\sin\left(\frac{2\pi}{n}\right) $$ and by the Squeeze Theorem, we get $$ \lim_{n\to\infty}n\sin(2\pi n!e)=2\pi $$

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One doubt that i have is that why are we not taking account of the integer that we get ? as $n\to \infty$ that can contribute a lot as well right ? so what am i still missing – Theorem Jul 27 '12 at 21:52
@Theorem: $\sin(2\pi n+2\pi x)=\sin(2\pi x)$ for any integer $n$. – robjohn Jul 27 '12 at 21:54
Very silly of me. Thanks. – Theorem Jul 27 '12 at 21:55
@robjohn: very detailed! (+1) – user 1618033 Aug 1 '12 at 13:19

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