Show the divergence of the series $\sum \frac{\sin^2n}{n}$ without Dirichlet's test 
Show that the series
  $$
\sum_{n\in\mathbb N} \frac{\sin^2n}{n}
$$
  is divergent. 

I know how to do this with the Dirichlet's test. But is there any other way to prove it? Thanks!
 A: Note that
$$\sum_{n=1}^{\infty} \frac{\sin^2(n)}{n} = \sum_{n=1}^{\infty} \frac{1}{2n} - \sum_{n=1}^{\infty} \frac{\cos(2n)}{2n}$$
which diverges since
$\sum \frac{1}{2n}$ diverges
$\sum \frac{\cos(2n)}{2n}$ converges
indeed by Abel transformation and Lagrange's trigonometric identities, let 
$$a_n=\frac{1}{2n} \quad b_n=\cos(2n)=B_{n}-B_{n-1}\quad B_n=\sum_{k=1}^{n} \cos (2k)= -\frac12+\frac{\sin(2n+1)}{2\sin 1}$$
$$S_N=\sum_{n=1}^{N}a_nb_n=\sum_{n=1}^{N} \frac{\cos(2n)}{2n}
=\frac{1}{2N}\left(-\frac12+\frac{\sin(2N+1)}{2\sin 1}\right)-\sum_{n=1}^{N-1} \left[ \left(-\frac12+\frac{\sin(2n+1)}{2\sin 1}\right)\left(\frac{1}{2n+2}-\frac{1}{2n}\right)\right]=$$
$$=\frac{1}{2N}\left(-\frac12+\frac{\sin(2N+1)}{2\sin 1}\right)-\sum_{n=1}^{N-1} \frac{1}{4n(n+1)}+\sum_{n=1}^{N-1} \frac{\sin (2n+1)}{4n(n+1)\sin 1}$$
and
$$\sum_{n=1}^{\infty} \frac{\sin (2n+1)}{4n(n+1)\sin 1}$$
converges absolutely by comparison with $\sum \frac{1}{n^2}$.
A: For any $a, b \in \mathbb{R}^n$, we have
$$\begin{align}\sin^2(a)  + \sin^2(b) 
&= \frac{1-\cos(2a)}{2} + \frac{1-\cos(2b)}{2}
&= 1 - \frac12(\cos(2a)+\cos(2b))\\ = 1 - \cos(a+b)\cos(a-b)
\end{align}
$$
For any integer $k$, this leads to
$$\sin^2(2k-1) + \sin^2(2k) = 1 - \cos(4k-1)\cos(1) \ge 1-\cos(1) > 0$$
As a result, we can bound the partial sums of even number of terms from below as
$$\sum_{n=1}^{2p}\frac{\sin^2(n)}{n}
= \sum_{k=1}^{p}\left(\frac{\sin^2(2k-1)}{2k-1} + \frac{\sin^2(2k)}{2k}\right)
\ge \sum_{k=1}^p \frac{1 - \cos(1)}{2k} = \frac{1-\cos(1)}{2}H_p$$
Since the harmonic number $H_p$ diverges like $\log(p)$ for large $p$, the partial sums of even number of terms diverges to $\infty$. As a consequence, the
sum $\displaystyle\;\sum_{n=1}^\infty \frac{\sin^2(n)}{n}$ diverges.
A: If you march around the unit circle in strides of radian $1$, you find that $|\sin n|\gt\sqrt3/2$ at least once for every six steps.  Thus
$$\sum{\sin^2n\over n}\gt\sum{3\over4(6k)}=\sum{1\over8k}=\infty$$
(Ah, I see that Daniel Fischer gave essentially the same answer as a comment.)
