sequences and series - Convergence of the following sum

Does the following sum converge? $$\sum_{n=1}^{\infty}\frac{\sin^2(n)}{n}$$
I tried the ratio test and got that $\rho=0$ which means that the series converges absolutely. However, Mathematica and Wolfram Alpha do not give a result when trying to find its convergence. Am I wrong?

Answer

Yes, you are wrong. The ratio test is inconclusive, and the series diverges.

Note that there is some $\varepsilon > 0$ such that $\sin^2(n) + \sin^2(n+1) > \varepsilon$ for all $n$. This is because if $n$ is close to a multiple of $\pi$, $n+1$ will not be. Thus $$\frac{\sin^2(n)}{n} + \frac{\sin^2(n+1)}{n+1} \ge \frac{\varepsilon}{n+1}$$

and a comparison with the harmonic series shows that the series diverges.