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?
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.