The limit of $\sin(n!)$

It is known that $\lim\limits_{n\to\infty}\sin n$ does not exist.

$\lim\limits_{n\to\infty}\sin(n!)$ exists or not?

I think there is a potentially different answer if the functions use radians or degrees. I say this because trigonometric functions relate to the circle. A complete circle is a whole number of degrees, but a transcendental number of radians. Factorials, meanwhile, are whole numbers.

For the sine function in degrees, the answer is that the limit is zero. I can say this because for every $$n \ge 360$$, $$360$$ divides $$n!$$. And if $$360$$ divides the number, then the sine of that number is zero.

For the sine function that uses radians, I can't think how to prove it at the moment, but I suspect the function does not converge.

• Actually, if we're talking degrees, we can narrow it down even further to $n \ge 180$, because $\sin 180^{\circ} = \sin 360^{\circ} = 0$. And in fact, because $6! = 720$, as long as $n \ge 6$, $\sin (n!)^{\circ} = 0$. Feb 15 '14 at 20:13
• The usual mathematical definition of $\sin$ is in radian, though. Feb 15 '14 at 20:15
• In fact, $6! = 720 = 2\times360$, so if $n$ is in degrees, $\sin(n) = 0$ for $n\ge 6$.
– user88319
Feb 15 '14 at 20:15
• Yes, certainly, there is a much earlier factorial that works. I simply chose the most obvious one to avoid needless complexity. The question wasn't about where the limit reached zero. Also, I agree, the real definitions of these functions are in radians, but I felt an answer addressing degrees was appropriate since many people use degrees most of the time. Feb 15 '14 at 20:24
• I think this question may be corollary to showing whether n! mod pi converges. That may be an easier question. If it does converge, then for any d, there is some n such that if m > n, then m! - n! mod pi < d. Feb 15 '14 at 21:09