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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    $\begingroup$ 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. $\endgroup$ – hmakholm left over Monica Oct 26 '11 at 16:31
  • $\begingroup$ You go above and beyond, Zev! (re: your edit) $\endgroup$ – The Chaz 2.0 Oct 26 '11 at 16:36
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    $\begingroup$ Thanks! I figure I'll get that Copy Editor badge eventually :) $\endgroup$ – Zev Chonoles Oct 26 '11 at 16:39
  • $\begingroup$ @Zev: It is certainly a nice fix. But, the $\infty$ in the second equation line should be an $n$. $\endgroup$ – Joe Johnson 126 Oct 26 '11 at 16:40
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    $\begingroup$ 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 $$0=\lim_{a\to 0}\;\frac 0a = \lim_{a\to 0}\;\lim_{b\to 0}\;\frac ba = \lim_{x\to 0}\;\frac xx = \lim_{x\to 0}\;1 = 1$$ $\endgroup$ – hmakholm left over Monica Oct 26 '11 at 16:58

(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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  • $\begingroup$ 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 $n \sin(2\pi e n!) \rightarrow 2\pi$. $\endgroup$ – mjqxxxx Oct 26 '11 at 16:57
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    $\begingroup$ 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 $2\pi$. $\endgroup$ – Craig Oct 26 '11 at 16:58
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    $\begingroup$ I am sorry.. the infinity on the sum was a typo.. please have a look at the revised version. Thank you. $\endgroup$ – M. Amin Oct 26 '11 at 17:00
  • $\begingroup$ Whoops, missed the $n$. Thanks, guys. $\endgroup$ – Thomas Andrews Oct 26 '11 at 17:13
  • $\begingroup$ @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 $n$. $\endgroup$ – Thomas Andrews Oct 26 '11 at 17:18

For given $n\geq2$ one has $$e\cdot n!=n!\sum_{k=0}^\infty{1\over k!}=n!\left(\sum_{k=0}^n{1\over k!}+\sum_{k=n+1}^\infty{1\over k!}\right)=m_n+r_n$$ with $m_n\in{\mathbb Z}$ and $${1\over n+1}<r_n={1\over n+1}+{1\over (n+1)(n+2)}+\ldots<{1\over n}+{1\over n^2}+\ldots<{1\over n-1}\ .$$ Since $$a_n:=n\>\sin\left(2\pi\cdot e\cdot n!\right)=n\>\sin(2\pi r_n)=n\ \ 2\pi r_n\ {\sin(2\pi r_n)\over 2\pi r_n}$$ and $r_n\to 0$ it follows that $$\lim_{n\to\infty}a_n=2\pi\lim_{n\to\infty}\bigl(n\> r_n\bigr)=2\pi\ .$$

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  • $\begingroup$ Funny, I thought that @Thomas already clearly explained this more than 3 years ago? $\endgroup$ – Did Nov 16 '14 at 19:01
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    $\begingroup$ @Did: The question was reposted by Community, I guess. At any rate Thomas' answer is not very streamlined, to say the least. What's the point in unearthing bygone questions when it is considered undecent to propose fresh answers? $\endgroup$ – Christian Blatter Nov 16 '14 at 20:09
  • $\begingroup$ The other answer looks allright to me. "Bygone questions" are "unearthed" because they have no accepted answer, and this is most often because the asker failed to follow the rules of the site (this one commented that "The rewrite is great. Thanks."), not because posted answers would be deficient. $\endgroup$ – Did Nov 16 '14 at 20:25
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    $\begingroup$ @Christian Blatter made it look so simpler. $\endgroup$ – Mittal G Feb 22 '16 at 9:04
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    $\begingroup$ Yes, I think Christian's explanation is much clearer and short and beautiful. $\endgroup$ – DonAntonio Feb 22 '16 at 9:08

Let's add another solution, which never hurts.

$$\lim_{n\rightarrow \infty}n\cdot \sin\left(2\pi e\cdot n!\right)= \lim_{n\rightarrow \infty}n\cdot \sin \left(2\pi \left(\sum_{k=0}^{\infty} \frac{1}{k!}\right)n!\right)$$

Let's study now that series.

$$\sum_{k=0}^{\infty}\left(\frac{1}{k!}\right) n!= \left(\sum_{k=0}^n\frac{1}{k!}+\sum_{k=n+1}^{\infty}\frac{1}{k!}\right)n!=A+b_n$$

Where $A\in \mathbb{Z}$ and:

$$b_n=\frac{1}{n+1}+ o \left( \frac{1}{n} \right)$$

From this, we can say that:

$$\lim_{n\rightarrow \infty}n\cdot \sin(2\pi A +2\pi b_n) = \lim_{n\rightarrow \infty}n\cdot \sin(2\pi b_n)=\lim_{n\rightarrow \infty}n\cdot \sin \left(\frac{2\pi}{n+1}+o\left(\frac{1}{n}\right)\right)$$

And, by expanding it using Taylor formulas, we obtain that the limit is equal to:

$$\lim_{n\rightarrow \infty}n\cdot\frac{2\pi}{n+1}+o\left(\frac{1}{n}\right)=2\pi.$$

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