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Does the sequence $\sin(n!)$ diverge(converge)?

It seems the sequence diverges. I tried for a contradiction but with no success. Thanks for your cooperation.

Michael Hardy
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M.R. Yegan
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  • Does the sequence $\sin n$ diverge (converge)? – Jika May 27 '14 at 14:07
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    If $n!$ is in degrees (instead of radians), then $\sin(n!)=0$ for all $n\geq6$. – barak manos May 27 '14 at 14:12
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    This looks harder than the divergence of $\sin n$. For one, if $\sin n!$ diverges, it follows automatically that $\pi$ is irrational, which is not an obvious fact. With $\sin n$, on the other hand, it is quite easy to prove divergence without knowing that $\pi$ is irrational. – Dan Shved May 27 '14 at 14:25
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    Same question for $\cos(n!)$, with some interesting commentary and answers: http://math.stackexchange.com/questions/8690/is-there-a-limit-of-cos-n – Hugh Denoncourt May 27 '14 at 23:51
  • My suspicion is that this diverges but that that is not easy to show. (This assumes radians are used.) – Michael Hardy Aug 09 '17 at 18:38

1 Answers1

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Depends on whether the parameter of $\sin$ is in radians or degrees. If in degree, $n!$ becomes multiple of 360 and after that function value will be zero, for all value of $n$.

In radians this will not happen as $\pi$ is irrational.

Orca
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  • I've up-voted this answer, but perhaps one should be explicit in saying that it is incomplete. It doesn't say whether the proposed sequence converges. But that is probably a very hard problem. – Michael Hardy Aug 09 '17 at 18:37