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