Convergence of $\sum_{n=1}^\infty \frac{\sin^2(n)}{n}$

Does the series $$ \sum_{n=1}^\infty \frac{\sin^2(n)}{n} $$ converge?

I've tried to apply some tests, and I don't know how to bound the general term, so I must have missed something. Thanks in advance.

Solution 1:

Hint:

Each interval of the form $\bigl[k\pi+{\pi\over6}, (k+1)\pi-{\pi\over6}\bigr)$ contains an integer $n_k$. We then have, for each $k$, that ${\sin^2(n_k)\over n_k}\ge {(1/2)^2\over (k+1)\pi}$. Now use a comparison test to show your series diverges.

Solution 2:

It is divergent:

Write $$\sum \frac{\sin^2(n)}{n} = \sum \frac{1}{2n} - \sum \frac{\cos(2n)}{2n},$$ clearly at the right handed side, the first is divergent but the second converges.