How to prove that $\lim_{n\to \infty}\sin(n!)$ does not exists?

Question

The continued fraction of $\pi$ has not yet be known. I could not see the distribution of the $\{\frac{n!}{2\pi}\}$ where $\{x\}=x-\lfloor x \rfloor$. Is there any idea how to find two convergent subsequences with different limits?

Answer 1

This answer to a closely related problem implies that, while the limit almost surely does not exist, current mathematical knowledge is unlikely to be capable of proving it.

Answer 2

I think I can show that the only possible limit is zero. I'll show my ideas and hope that they might be useful.

If $L = \lim_{n \to \infty} \sin(n!) $ exists, then, for large enough $n$, $(n+1)! \approx n!+2k_n\pi $ where $k_n$ is an integer that depends on $n$.

Then $k_n \approx \frac1{2\pi}((n+1)!-n!) = \frac1{2\pi}nn! $.

Similarly, $k_{n+1} \approx \frac1{2\pi}((n+1)!-n!) = \frac1{2\pi}(n+1)(n+1)! $ so

$\begin{array}\\ k_{n+1}-k_n &\approx \frac1{2\pi}((n+1)(n+1)!-nn!)\\ &= \frac1{2\pi}n!((n+1)(n+1)-n)\\ &= \frac1{2\pi}n!(n^2+n+1)\\ &= \frac1{2\pi}n!(n^2+n)+\frac1{2\pi}n!\\ &= \frac1{2\pi}nn!(n+1)+\frac1{2\pi}n!\\ &\approx k_n(n+1)+\frac1{2\pi}n!\\ \end{array} $

so $\frac1{2\pi}n! \approx k_{n+1}-(n+2)k_n $.

Therefore $n!/(2\pi)$ is close to an integer, so $\sin(n!) \approx 0$.

Therefore the only possible limit is zero.

If we can choose $n$ so that the fractional part of $\frac1{2\pi}((n+1)!-n!) \approx \pi/2 $, then $\sin(n!)$ and $\sin((n+1)!)$ will not be close so the limit can not exist.

This last, of course, does not depend on the limit being zero.

I don't know where to go from here, so I'll stop.