Interesting Limit (15)

$e\cdot n! = \sum_{k=0}^\infty \frac{n!}{k!} = \left(\sum_{k=0}^n \frac{n!}{k!}\right) + \left(\sum_{k=n+1}^\infty \frac{n!}{k!}\right) := A_n + B_n$

Here, we note that $A_n$ is an integer, so $\sin(2\pi en!) = \sin(2\pi A_n+ 2\pi B_n) = \sin(2\pi B_n),$ and $\left|(n+1) B_n - 1 \right| = \sum_{k=n+2}^\infty \frac{(n+1)!}{k!} \leq \sum_{k=n+2}^\infty \frac{1}{n^{k - (n+1)}} = \frac{\frac{1}{n}}{1-\frac{1}{n}} = \frac{1}{n-1}\xrightarrow{n\to\infty} 0$ so that $\lim_{n\to\infty} n B_n = \lim_{n\to\infty} \frac{n}{n+1} \cdot (n+1) B_n = 1 \cdot 1 = 1.$

Therefore $\lim_{n\to\infty} n \sin(2\pi en!) = 2\pi \lim_{n\to\infty} \left(n B_n\right) \left(\frac{\sin(2\pi B_n)}{2\pi B_n}\right) = 2\pi(1)(1) = 2\pi \approx 6.28$

---

For the Bonus question:

Assume $e$ is rational, so $e = \frac{M}{N}$ for some integers $M,N$ with $N > 0$ . However, then for $n \geq N$ , we'd have $e n! = M \cdot \frac{n!}{N}$ is an integer and therefore $\sin(2\pi en!) = 0\text{.}$ It would then follow that the limit in the question is also $0\text{.}$ Since it isn't, $e$ must be irrational.

We write

$\displaystyle n!=\int_{0}^{\infty} x^{n}e^{-x}dx$

So,

$\displaystyle en!=\int_{0}^{\infty} x^{n}e^{-(x-1)}dx=\int_{0}^{\infty} (1+u)^{n}e^{-u}du+\int_{-1}^{0} (1+u)^{n}e^{-u}du$

The first term is clearly an integer and the second term tends to 0 as n tends to infinity.

So, the limit can be written in the form

$\displaystyle \lim_{n->\infty} n\sin{(2\pi e n!)}=\lim_{n->\infty} 2\pi n \int_{-1}^{0} (1+u)^{n}e^{-u}du$

And by integrating by parts,

$\displaystyle \int_{-1}^{0} (1+u)^{n}e^{-u}du=\dfrac{1}{n+1}+\dfrac{1}{n+1} \int_{-1}^{0} (1+u)^{n+1}e^{-u}du$

Hence,

$\displaystyle \lim_{n->\infty} n \int_{-1}^{0} (1+u)^{n}e^{-u}du=\lim_{n->\infty} \dfrac{n}{n+1}+\dfrac{n}{n+1} \int_{-1}^{0} (1+u)^{n+1}e^{-u}du=1$

So, the answer is $\displaystyle \boxed{2\pi}$

Strangely enough I assumed this was a typo and the argument in the sin was devided by n! When you apply the sterling approximation you arrive at the same result:)