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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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  • $\begingroup$ 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. $\endgroup$ Oct 26, 2011 at 16:53
  • $\begingroup$ 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? $\endgroup$
    – M. Amin
    Oct 26, 2011 at 16:53
  • $\begingroup$ @M.Amin: Didn't you intend for the second line to have an $n$ in the upper limit of the sum, i.e. $$ n!\sum_{k=0}^n\frac{1}{k!}=\frac{n!}{0!}+\frac{n!}{1!}+\cdots+\frac{n!}{n!}=m $$ $\endgroup$ Oct 26, 2011 at 16:54
  • $\begingroup$ @HenningMakholm: So expressing e itself as a limit inside the bigger limit process is not valid? $\endgroup$
    – M. Amin
    Oct 26, 2011 at 16:55
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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 $$0=\lim_{a\to 0}\;\frac 0a = \lim_{a\to 0}\;\lim_{b\to 0}\;\frac ba = \lim_{x\to 0}\;\frac xx = \lim_{x\to 0}\;1 = 1$$ $\endgroup$ Oct 26, 2011 at 16:58

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 $2\pi$.

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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, 2011 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, 2011 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, 2011 at 17:00
  • $\begingroup$ Whoops, missed the $n$. Thanks, guys. $\endgroup$ Oct 26, 2011 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 $n$. $\endgroup$ Oct 26, 2011 at 17:18
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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, 2014 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? $\endgroup$ Nov 16, 2014 at 20:09
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    $\begingroup$ @Christian Blatter made it look so simpler. $\endgroup$
    – PAMG
    Feb 22, 2016 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, 2016 at 9:08
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    $\begingroup$ In your bound for $r_n$, you should have $\sum_{i=1}^\infty n^{-i} = (n-1)^{-1}$ rather than $\sum_{i=1}^\infty n^{-i} < (n-1)^{-1}$. $\endgroup$
    – mlbaker
    Aug 14, 2020 at 0:12
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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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