Find the following limits:
(1) $\limsup_{n\to\infty } \sin (n!) $
(2) $\liminf_{n\to\infty } \sin (n!) $
(3) $\limsup_{n\to\infty } \cos (n!) $
(4) $\liminf_{n\to\infty } \cos (n!) .$
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3$\begingroup$ MO is for questions of math research. $\endgroup$– Gerry MyersonDec 4, 2014 at 5:07
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1$\begingroup$ This is an interesting question because it looks less hopeless than many other seemingly similar questions. This one provides a chance for solving it. $\endgroup$– Włodzimierz HolsztyńskiDec 4, 2014 at 5:24
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$\begingroup$ I'd simply ask about the behavior of $\ \exp(i\!\cdot\! n!)$ $\endgroup$– Włodzimierz HolsztyńskiDec 4, 2014 at 5:25
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1 Answer
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My guess is that the lim infs are $-1$ and the lim sups are $+1$, but I think this requires better information about rational approximations of $\pi$ than we have available. What I can say, though, is that $\lim_{n \to \infty} \sin(n! \pi e) = 0$, while $\lim_{n \to \infty} \cos((2n)! \pi e) = -1$ and $\lim_{n \to \infty} \cos((2n+1)! \pi e) = +1$.
Now we strongly suspect, but haven't been able to prove, that $\pi e$ is irrational. If by some miracle it happens to be rational, then the limit points of $\exp(n! i)$ as $n \to \infty$ are a finite set of roots of unity.
answered Dec 4, 2014 at 5:14