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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?

  • 2
    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.2011-10-26
  • 0
    You go above and beyond, Zev! (re: your edit)2011-10-26
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    Thanks! I figure I'll get that Copy Editor badge eventually :)2011-10-26
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    @Zev: It is certainly a nice fix. But, the $\infty$ in the second equation line should be an $n$.2011-10-26
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    How did you put $n! e=m$ it is not precise!2011-10-26
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    @Joe: Thanks for catching it, I've fixed it now.2011-10-26
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    @Ali: That mistake was my fault.2011-10-26
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    @HenningMakholm I see it typed correctly.. not as you see them. I typed them in Google Docs. Maybe there's something wrong at your end? I don't really know how to use MathJax..2011-10-26
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    This calculation is based on letting $a\to e$ in $n\sin(2\pi a n!)$, while _synchronously_ letting $n\to\infty$. Coalescing limits in this way is not generally valid. At most you can conclude that _if the limit exists_ it will be $0$. But I'm not even sure of that.2011-10-26
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    Ok I can see that @ZevChonoles fixed my post.. thanks Zev.. I also fixed the little mistake about n and infinity.. so.. can anyone check my result tell me is right or wrong?2011-10-26
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    @M.Amin: Didn't you intend for the second line to have an $n$ in the upper limit of the sum, i.e. $$ n!\sum_{k=0}^n\frac{1}{k!}=\frac{n!}{0!}+\frac{n!}{1!}+\cdots+\frac{n!}{n!}=m $$2011-10-26
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    @HenningMakholm: So expressing e itself as a limit inside the bigger limit process is not valid?2011-10-26
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    @Zev: Yes that's how I want it yea..2011-10-26
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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 $$0=\lim_{a\to 0}\;\frac 0a = \lim_{a\to 0}\;\lim_{b\to 0}\;\frac ba = \lim_{x\to 0}\;\frac xx = \lim_{x\to 0}\;1 = 1$$2011-10-26
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    I fixed all the typos now2011-10-26
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    @Henning: But I don't see how lima→0limb→0b/a is equal to limx→0x/x..2011-10-26
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    @M.Amin, it _isn't_. That's the point: collapsing the two limits is not valid, but that is what you were doing in your calculation.2011-10-26
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    @Henning: Ok I get your point. But could you refer me to some text concerning this?2011-10-26
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    @M. Amin: I think you meant $n!\sum_{k=0}^{n}\frac{1}{k!} $ is an integer. is that right?2011-10-26
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    @Gardel: Yes that's what I meant.2011-10-26
  • 0
    This might be helpful : fa.its.tudelft.nl/~teuwen/wip/Limit.pdf2014-11-16

3 Answers 3

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(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 $2\pi$.

  • 0
    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 $n \sin(2\pi e n!) \rightarrow 2\pi$.2011-10-26
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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 $2\pi$.2011-10-26
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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.2011-10-26
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    Whoops, missed the $n$. Thanks, guys.2011-10-26
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    @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 $n$.2011-10-26
  • 0
    Thanks a lot that was helpful. But I think in your third equation, it should be n^(n-m) not n^(m-n), right?2011-10-26
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    @Amin, yes. Fixed2011-10-26
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    The rewrite is great. Thanks.2011-10-26
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    Thanks Thomas Andrews for nice solution.2013-11-05
35

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 $${1\over n+1}\sin\left(2\pi\cdot e\cdot n!\right)=n\>\sin(2\pi r_n)=n\ \ 2\pi r_n\ {\sin(2\pi r_n)\over 2\pi r_n}$$ 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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    Funny, I thought that @Thomas already clearly explained this more than 3 years ago?2014-11-16
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    @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?2014-11-16
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    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.2014-11-16
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    @Christian Blatter made it look so simpler.2016-02-22
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    Yes, I think Christian's explanation is much clearer and short and beautiful.2016-02-22
4

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:

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