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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?

  • 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
    Thanks Thomas Andrews for nice solution.2013-11-05
38

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} 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 $\lim_{n\to\infty}a_n=2\pi\lim_{n\to\infty}\bigl(n\> r_n\bigr)=2\pi\ .$

  • 4
    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:

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