Series $\sum \frac{1}{n^2\sin^3n}$

Question

Question : Show that series $\sum \cfrac{1}{n^{2}\sin^{3}n}$ is divergent.

Hint:

Show that $$\sum \frac{1}{n|\sin(n)|}$$ is divergent.

I am interested in other possible proofs for this question.

Answer 1

Since $\pi$ is irrational, by Hurwitz's theorem, there are infinitely many pairs of relative prime integers $(n,m)$ such that

$$\left|\pi - \frac{n}{m}\right| < \frac{1}{\sqrt{5}{m^2}} \quad\implies\quad|n - m\pi| < \frac{1}{\sqrt{5}m} $$ We can assume both $n, m > 0$ and for any such pair, we have $$ |\sin n | = |\sin(n-m\pi)| < \sin\frac{1}{\sqrt{5}m} < \frac{1}{\sqrt{5}m} = \frac{1}{\sqrt{5}n}\left( \pi + ( \frac{n}{m} - \pi )\right) < \frac{1}{\sqrt{5}n}\left( \pi + 1\right) \\ \implies \frac{1}{n^2 |\sin n|^3} > \left(\frac{\sqrt{5}}{\pi+1}\right)^3 n$$ This means the sequence $\displaystyle\;\frac{1}{n^2\sin^{3}n}\;$ doesn't converge to $0$ and hence the series diverges.