# What is the limit of $n \sin (2 \pi \cdot e \cdot n!)$ as $n$ goes to infinity?

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?

• 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. Oct 26 '11 at 16:53
• 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? Oct 26 '11 at 16:53
• @M.Amin: Didn't you intend for the second line to have an $n$ in the upper limit of the sum, i.e. $$n!\sum_{k=0}^n\frac{1}{k!}=\frac{n!}{0!}+\frac{n!}{1!}+\cdots+\frac{n!}{n!}=m$$ Oct 26 '11 at 16:54
• @HenningMakholm: So expressing e itself as a limit inside the bigger limit process is not valid? Oct 26 '11 at 16:55
• 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 $$0=\lim_{a\to 0}\;\frac 0a = \lim_{a\to 0}\;\lim_{b\to 0}\;\frac ba = \lim_{x\to 0}\;\frac xx = \lim_{x\to 0}\;1 = 1$$ 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 $2\pi$.

• 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 $n \sin(2\pi e n!) \rightarrow 2\pi$. Oct 26 '11 at 16:57
• 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 $2\pi$. Oct 26 '11 at 16:58
• I am sorry.. the infinity on the sum was a typo.. please have a look at the revised version. Thank you. Oct 26 '11 at 17:00
• Whoops, missed the $n$. Thanks, guys. 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 $n$. 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 $${1\over n+1} Since $$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\ .$$

• Funny, I thought that @Thomas already clearly explained this more than 3 years ago?
– Did
Nov 16 '14 at 19:01
• @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? Nov 16 '14 at 20:09
• @Christian Blatter made it look so simpler. Feb 22 '16 at 9:04
• Yes, I think Christian's explanation is much clearer and short and beautiful. Feb 22 '16 at 9:08
• In your bound for $r_n$, you should have $\sum_{i=1}^\infty n^{-i} = (n-1)^{-1}$ rather than $\sum_{i=1}^\infty n^{-i} < (n-1)^{-1}$. Aug 14 '20 at 0:12

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:

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