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Calculate the limit

$$ \lim_{n\rightarrow\infty}(\sin(n!)+1)^{1/n} $$

or prove that the limit does not exist.

This appeared as a problem in my mathematical analysis test, and the answer was that the limit exists and it was $1$. But later the teacher found a mistake in his proof and eventually removed the problem from the test. But I'm just curious. Does this problem have a certain answer?

The biggest question for me, is that I can't show that there does not exist any $n_0$ so that $\sin(n_0!)$ is close to $-1$ enough so that the original term might not converge. Any help would be appreciated!

Giorgos Giapitzakis
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Mivik
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  • Here is a similar question about $\cos(n!)$. The answer (as I understand it) is “we don't know.” – Martin R Nov 15 '22 at 15:51
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    @MartinR we know $(\sin(n))_n$ is equidistributed, but do we know $(\sin(n!))_n$ is equidistributed? If it isn't, then we know for sure that the limit is $1$ – FShrike Nov 15 '22 at 16:55
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    @FShrike: The problem of the density or lacunarity of the limit points of the sequence $\sin n!$ is probably as hard as that of $\cos(n!)$. – Mittens Nov 15 '22 at 18:18
  • @FShrike Please can you justify your claim? It seems wrong, or at best, unjustified to me. For example, it's not clear to me why we cannot have: $\ (\sin(n!))n\ $ not equi-distributed and $\liminf{n\rightarrow\infty}(\sin(n!)+1)^{1/n}<1.$ – Adam Rubinson Nov 15 '22 at 20:47
  • I was being sloppy with my language, I meant one thing and wrote another @AdamRubinson . I was thinking: "... if $\sin(n!)$ is bounded away from $-1$, then..." but although: equidistributed --> not bounded away from $-1$, the converse fails! You are quite right, it's unjustified at present – FShrike Nov 15 '22 at 21:25
  • Related https://math.stackexchange.com/questions/677719/the-limit-of-sinn?noredirect=1&lq=1 and https://mathoverflow.net/questions/45665/distribution-mod-1-of-factorial-multiples-of-real-numbers – Enrico M. Nov 15 '22 at 22:54

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