How can I prove that $a_n>0$ infinitely often? Let $n\in\mathbb{N}$ and $$a_n:=sin(2\pi^2(2n+1)!)$$
How can I prove that $a_n>0$ infinitely often?
Clearly,   $a_n>0$ infinitely often is equivalent to {$\pi(2n+1)!$}$\leq 0.5$ infinitely often where {.} is fractional part function. Since $\pi$ is irrational, {$\pi(2n+1)!$} will never be zero. How can we show that {$\pi(2n+1)!$}$\leq 0.5$ infinitely many times? Graph of $y=a_n$ is available here (Desmos). On observing the graph, it seems that $a_n>0$ is true infinitely often. I  tried method of induction etc.  but I am not able to prove this. What are the various ways to prove it and how can it be proved?
 A: To elaborate on previous answers and comments, you need to prove that there are infinitely many pairs $(k,n)$ such that
$$
\left|(2n+1)! - \frac{k}{\pi}\right| < \frac{1}{4\pi}.
$$
Phrased in the language of Diophantine approximation, one needs to find infinitely many rational approximations to $\frac{1}{\pi}$ of the form $\frac{(2n+1)!}{k}$; i.e., infinitely many pairs $(k,n)$ such that
$$
\left|\frac{(2n+1)!}{k} - \frac{1}{\pi}\right| < \frac{1}{4\pi k}.
$$
It is highly nontrivial to claim that an infinity of these pairs exists. Even to say that there are infinitely many rationals $\frac{n}{k}$ such that
$$
\left|\frac{n}{k} - \frac{1}{\pi}\right| < \frac{1}{4\pi k}.
$$
requires a nontrivial application Dirichlet's approximation theorem. For this to still hold when $n$ is restricted to the odd factorials, one needs to show that the factorials don't "conspire" in some way so as to produce poor rational approximations to $\frac{1}{\pi}$. It certainly seems like this shouldn't happen, but actually giving a rigorous proof (in either direction) is nontrivial.
Terry Tao has a good article on "conspiracies" among prime numbers. In general, proving that conspiracies don't exist in the integers is a very difficult task.
