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?
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Hint: $-1\le \sin(n!) \le 1$ – Learning Apr 09 '18 at 04:33
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3Duplicate of a question that was never adequately answered in the first place. – CogitoErgoCogitoSum Apr 09 '18 at 04:37
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I think the best bet here is to use the Cauchy Convergence criterion. – CogitoErgoCogitoSum Apr 09 '18 at 04:41
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1@PrithiviRaj: I'm pretty sure the idea you had when you posted that hint is missing the main problem here. – Apr 09 '18 at 04:45
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@Hurkyl Hmmm... I dont follow the primary answer, sadly. And it isnt even the same question, but close. Im still leaning toward the Cauchy convergence. Pick the right trig identities and the right choices of m,n>N, it should simplify... maybe? – CogitoErgoCogitoSum Apr 09 '18 at 04:52
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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.
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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.
marty cohen
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There is a complication here; this argument attacks the issue of whether $n!$ converges to a point in the circle $\mathbb{R} / 2 \pi$. The limit of $\sin(n!)$, unfortunately can exist in more general situations, such as if $n!$ bounces between two angles with the same sine. – Apr 09 '18 at 05:16
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A more serious issue lies with the claim $\frac{1}{2\pi} n n! (n+1)$ should approximate $k_n (n+1)$; why must the error between $k_n$ and $\frac{1}{2 \pi} n n!$ be asymptotically $o(\frac{1}{n})$? – Apr 09 '18 at 05:25
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1I know there are lots of problems with my attempt. I just wanted to put some ideas out there. – marty cohen Apr 09 '18 at 10:35