# Divergence of the sequence $\sin(n!)$

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.

• Does the sequence $\sin n$ diverge (converge)? – Jika May 27 '14 at 14:07
• If $n!$ is in degrees (instead of radians), then $\sin(n!)=0$ for all $n\geq6$. – barak manos May 27 '14 at 14:12
• 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
• Same question for $\cos(n!)$, with some interesting commentary and answers: 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 9 '17 at 18:38

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.
Hint: Take a subsequence in which $a_{n_i} \approx \pi ( 4i+1)/2$ and another where $b_{n_j} \approx \pi ( 4i +3)/2$.
• Your notation is confused; presumably you mean something like $n_i!\to\pi/2\pmod{2\pi}$. But more importantly, are you sure that your hint can actually be followed through? According to this discussion on MathOverflow, the divergence of $\cos(n!)$, which is equivalent to the divergence of $\sin(n!)$, would imply that $e+\frac{1}{2\pi}$ is irrational, solving an open problem. Your hint works for $\sin n$, but it seems like no one knows whether such subsequences exist here. – epimorphic Feb 20 '15 at 2:52
• My hint is to realize you may choose a subsequence in which $sin(n_i!) > 0$ and another in which $sin(n_i)<0$. I guess my notation was bad, but I meant approximately around $1$ and $-1$ respectively. – Jeb Feb 20 '15 at 15:23
• As I wrote after that, the "may choose" part needs to be justified because no one has ever been able to prove the simultaneous existence of both sequences. We can't just appeal to the irrationality of $\pi$ as one does when proving that $\sin n$ diverges. For example, one of the other answers in the MO question I linked to shows that $n! \to 0 \pmod{e}$. – epimorphic Feb 20 '15 at 15:31