How to prove that the series $\sum\limits_{n=1}^\infty {\sin^2n} $ diverges I want to use a divergence test to prove that $\lim_{n\to \infty} \sin^2n$ does not converge. So $\sum_{i=1}^\infty \sin^2 n $ diverge. But because $\pi$ is an irrational number. So I cannot use subsequence with $n=n\pi$ and $n=\frac{(2n-1)\pi}{2}$. So how can I prove that the sum diverges?
Where $n \in \Bbb{N}$

Lemma. If $\lvert \sin x\rvert<1/4$, then $\lvert \sin (x+1)\rvert>1/4$.
Proof. Observe that $\frac{\pi}{4}<1<\frac{\pi}{3}$, and hence $\sin 1>\frac{\sqrt{2}}{2}$.
Also, $\lvert \cos x\rvert>\sqrt{1-\left(\frac{1}{4}\right)^2}=\sqrt{\frac{15}{16}}$. Hence
$$
\lvert \sin(x+1)\rvert\ge \lvert \sin 1\cos x\rvert-\lvert \sin x\cos 1\rvert \\
\ge \frac{\sqrt{2}}{2}\frac{\sqrt{15}}{4}-\frac{1}{4}=\frac{\sqrt{30}}{8}-\frac{1}{4}>\frac{1}{2}-\frac{1}{4}=\frac{1}{4} 
.
$$
This implies that $\{\sin n\}$ DOES NOT converge to zero, since if $\sin n\to 0$, then eventually $\lvert \sin n\rvert<1/4$, for all $n\ge n_0$, for some $n_0\in\mathbb N$. But the Lemma above proves that this is impossible!
Now that means $\sin n\not\to 0$, then also $\sin^2 n\not\to 0$, and hence
the series $\sum \sin^2 n$ cannot converge.
