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Does the sequence $\sin(n!\pi^2)$ converge or diverge?

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    $\begingroup$ If $\pi^2$ were rational, then the sequence would certainly converge to $0$, but as it is, I suspect this may be a hard problem. I would guess that $\sin(n!\pi^2)$ oscillates erratically with limsup and liminf respectively $1$ and $-1$, but I think proving it might take some work. $\endgroup$ – Michael Hardy Mar 10 '14 at 18:59
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    $\begingroup$ @MichaelHardy Ok, now I feel stupid, I don't see it. Why does $\sin(n!q)$ necessarily converge to 0 for $q \in \mathbb{Q}$? $\endgroup$ – fgp Mar 10 '14 at 19:04
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    $\begingroup$ I believe this is equivalent to an unknown one, $$ n! \pi - \lfloor n! \pi \rfloor $$ $\endgroup$ – Will Jagy Mar 10 '14 at 19:06
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    $\begingroup$ @MichaelHardy : I think you forgot a $\pi$ in your reasoning... sinus of an integer is never zero unless you have $\sin(0)$ which is not the case here. $\endgroup$ – user88595 Mar 10 '14 at 19:13
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    $\begingroup$ @JacobWakem : Any reason (or intuition) why it would converge to an integer? $\endgroup$ – user88595 Mar 10 '14 at 19:49
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I devoted a chapter (3) of my thesis on the set $\displaystyle G = \{x \in \mathbb{R}: \lim_{n \rightarrow \infty} \sin{(n! \pi x) = 0}\}$. It is easy to see that (Euler's constant) $e \in G$. Let alone $\pi$, I do not even know if $e^2 \in G$.

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I cant answer the question but here are some thoughts. $\sin(n \pi^2)$ does not have a limit because $\{ n\pi \}$ is dense in $[0,1]$. So the question would really be is $\{ n! \pi \}$ is dense. One might ask if $\{ n! \alpha \}$ is dense for any irrational $\alpha$, but it is easy to see that $\{ n! e \}\rightarrow 0$ from the series definition of $e$. So $\sin(n! e \pi)$ limits to $0$. Thus the behaviour of $\{ n! \pi \}$ depends on properties of $\pi$.

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