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I tried and got this

$$e=\sum_{k=0}^\infty\frac{1}{k!}=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}$$ $$n!\sum_{k=0}^n\frac{1}{k!}=\frac{n!}{0!}+\frac{n!}{1!}+\cdots+\frac{n!}{n!}=m$$ where $m$ is an integer. $$\lim_{n\to\infty}n\sin(2\pi en!)=\lim_{n\to\infty}n\sin\left(2\pi n!\sum_{k=0}^n\frac{1}{k!}\right)=\lim_{n\to\infty}n\sin(2\pi m)=\lim_{n\to\infty}n\cdot0=0$$

Is it correct?

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What does the notation ${}^n\sin^{(2\pi en!)}$ mean for you? If you mean $n\sin(2\pi e n!)$, then you might as well just type your calculations into the question as best you can, rather than link to an external rendering that is badly typeset anyway. –  Henning Makholm Oct 26 '11 at 16:31
    
You go above and beyond, Zev! (re: your edit) –  The Chaz 2.0 Oct 26 '11 at 16:36
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Thanks! I figure I'll get that Copy Editor badge eventually :) –  Zev Chonoles Oct 26 '11 at 16:39
    
@Zev: It is certainly a nice fix. But, the $\infty$ in the second equation line should be an $n$. –  Joe Johnson 126 Oct 26 '11 at 16:40
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Yes, expressing $e$ as a limit is valid in any context. What is not valid is to combine the two limits and doing them in one operation. Otherwise you could prove $0=1$ by reasoning $$0=\lim_{a\to 0}\;\frac 0a = \lim_{a\to 0}\;\lim_{b\to 0}\;\frac ba = \lim_{x\to 0}\;\frac xx = \lim_{x\to 0}\;1 = 1$$ –  Henning Makholm Oct 26 '11 at 16:58
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1 Answer

(Added fix recommended by Craig in comments, and complete rewrite for clarity.)

We will use the following: $\lim_{x\rightarrow 0} {\frac{\sin x}{x}}=1$.

Lemma: If $\{x_n\}$ is a sequence (of non-zero values) that converges to $0$, then $$\lim_{n\rightarrow\infty}{n \sin{x_n}} = \lim_{n\rightarrow\infty} nx_n$$

Proof: Rewrite $n\sin{x_n} = n x_n \frac{\sin{x_n}}{x_n}$. The lemma follows since $\sin{x_n}/x_n \rightarrow 1$ by above.

Now, let $[[x]]$ be the fractional part of $x$. Let $e_n = [[n!e]]$.

Lemma: For $n>1$, $e_n\in (\frac{1}{n+1}, \frac{1}{n-1})$

Proof: $$n!e = K + \sum_{m=n+1}^\infty \frac{n!}{m!}$$

Where $K$ is an integer.

But for $m>n$, $\frac{n!}{m!} = \frac{1}{(n+1)(n+2)...m} < n^{n-m}$.

So $$\frac{1}{n+1}<\sum_{m=n+1}^\infty \frac{n!}{m!} < \sum_{m=n+1}^\infty n^{n-m} = \sum_{k=1}^\infty n^{-k}$$

But the right hand side is a geometric series whose sum is $\frac{1}{n-1}$.

So $n!e-K\in(\frac{1}{n+1}, \frac{1}{n-1})$, and, since $K$ is an integer, it must be $e_n=n!e-K$.

Theorem: $\lim_{n\rightarrow \infty} n \sin(2\pi n! e) = 2\pi$

Proof: By periodicity of $\sin$, $\sin(2\pi n! e) = \sin(2\pi e_n)$.

Letting $x_n = 2\pi e_n$, we see, from our first lemma:

$$\lim n \sin x_n = \lim n x_n$$

But $nx_n = 2\pi ne_n$, and, since $ne_n\in(\frac{n}{n+1},\frac{n}{n-1})$, we see that $ne_n\rightarrow 1$. So our limit is $2\pi$.

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The question is about the behavior of $n\sin(2\pi e n!)$, though. So you need to go one step further: you've shown that $e n! \approx K + 1/n + O(1/n^2)$; so $\sin(2\pi e n!) \approx 2\pi/n + O(1/n^2)$; and finally $n \sin(2\pi e n!) \rightarrow 2\pi$. –  mjqxxxx Oct 26 '11 at 16:57
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He asked for the limit of $n \sin (2\pi n! e)$, not $\sin (2\pi n! e)$. However, you've shown that the fractional part of $n! e$ is in $(1/n, 1/(n-1))$, so the limit is $2\pi$. –  Craig Oct 26 '11 at 16:58
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I am sorry.. the infinity on the sum was a typo.. please have a look at the revised version. Thank you. –  M. Amin Oct 26 '11 at 17:00
    
Whoops, missed the $n$. Thanks, guys. –  Thomas Andrews Oct 26 '11 at 17:13
    
@Craig, I've actually only shown it is in $(\frac{1}{n+1},\frac{1}{n-1})$. But yes, the limit is $2\pi$ with the added factor of $n$. –  Thomas Andrews Oct 26 '11 at 17:18
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