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 $n \in \Bbb{N}$
 A: 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.
A: It's something like you suggested. Consider arbitrary $n$. Then if $|sin(n)|<{1\over\sqrt{2}}$ then $n \in (\pi k -{\pi \over 4}, \pi k + {\pi \over 4})$ for some $k$. Hence $n+1$ or $n+2$ lying in interval $(\pi k + {\pi \over 4}, \pi k + {3\pi \over 4})$ and $max\{|\sin(n+1)|,|\sin(n+2)|\} \ge {1 \over \sqrt{2}}$. We just showed that for every $n$ $\max\{\sin^2(n),\sin^2(n+1),\sin^2(n+2)\}\ge{1\over2}$. 
So there is sequence $\{n_k\}$ such that $\sin^2(n_k) \ge {1\over2}$ and $n_k \to \infty$. Obviously $\sum\limits_{n=1}^{\infty} sin^2(n) \ge \sum\limits_{k=1}^{\infty} sin^2(n_k) \ge \sum\limits_{n=1}^{\infty} {1\over2} = \infty$
A: Consider the following integer sequence:
$$
m_k = \left\lfloor\frac{\pi}{2}\right\rfloor +\left\lfloor2\pi\cdot 10^k\right\rfloor.
$$
One may see that as 
$$
\lim_{k\to\infty}\left(2\pi\cdot 10^k - \left\lfloor2\pi\cdot 10^k\right\rfloor\right) = 0$$
we get that
$$
\lim_{k\to\infty}\left(\frac{\pi}{2} + 2\pi\cdot 10^k - m_k\right) = \frac{\pi}{2} - \left\lfloor\frac{\pi}{2}\right\rfloor
$$
which follows
$$
\lim_{k\to\infty}\sin\left(\frac{\pi}{2} + 2\pi\cdot 10^k - m_k\right) = \sin\left(\frac{\pi}{2} - \left\lfloor\frac{\pi}{2}\right\rfloor\right)
$$
or
$$
\lim_{k\to\infty}\cos m_k = \cos\left\lfloor\frac{\pi}{2}\right\rfloor \Rightarrow \lim_{k\to\infty}\sin m_k = \sin\left\lfloor\frac{\pi}{2}\right\rfloor > 0.
$$
Thus subsequence $\sin(m_k)$ converges to nonzero number and so does subsequence $\sin^2(m_k)$ so the whole sum diverges. 
A: If $\{\sin^2n\}$ converges to $0$, then so does $\{\sin n\}$. But $\{\sin n\}$ diverges as shown in this MSE question.
