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

M.R. Yegan
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  • What do you thin? And why? – fgp Mar 10 '14 at 18:53
  • As $n$ increases? Did you try with values to see what happens? – user88595 Mar 10 '14 at 18:53
  • Also, do you think $\pi^2$ is important? Did you try with another value instead? – user88595 Mar 10 '14 at 18:54
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    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. – Michael Hardy Mar 10 '14 at 18:59
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    @MichaelHardy Ok, now I feel stupid, I don't see it. Why does $\sin(n!q)$ necessarily converge to 0 for $q \in \mathbb{Q}$? – fgp Mar 10 '14 at 19:04
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    I believe this is equivalent to an unknown one, $$ n! \pi - \lfloor n! \pi \rfloor $$ – Will Jagy Mar 10 '14 at 19:06
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    @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. – user88595 Mar 10 '14 at 19:13
  • Converges. Let n>2 . sin(n!π 2 )=sin(2π(n!π/2))=sin(0) (period of sin is 2π ). This doesn't work because n!/pi/2 is not an integer. But if this converges to being an integer we are done... Thus, if (iff?) Jagy's expression converges to 0, we have the desired result. – Jacob Wakem Mar 10 '14 at 19:27
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    @JacobWakem : Any reason (or intuition) why it would converge to an integer? – user88595 Mar 10 '14 at 19:49
  • @user88595 : no; I just believe that is the crux of the problem. – Jacob Wakem Mar 10 '14 at 20:12
  • OK, I should have said if $\pi$ were rational then $\sin(n!\pi^2)$ would converge to $0$. The point is that if $\pi=a/b$, and $a$ and $b$ are integers, then $n!\pi^2=(n!\pi)\pi$ $=(n!a/b)\pi$, and if $n\ge b$, then $n!/b$ is an integer and so is $n!a/b$, so $(n!a/b)\pi$ is an integer multiple of $\pi$, so that its sine is $0$. – Michael Hardy Mar 11 '14 at 04:13
  • @MichaelHardy: I get what you wanted to say but then you have a multiple of $\pi$ which you defined to be rational, sin of it won't be zero. You are splitting $\pi^2$ into $\pi\times\pi$, one you say is a rational, the other one you say isn't... – user88595 Mar 11 '14 at 09:47

2 Answers2

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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 $e^2 \in G$.

hot_queen
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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 $\pi$.

Rene Schipperus
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