Calculus (3rd Edition)

The series $\Sigma_{n=1}^{\infty}\frac{\cos^2 n}{n^2}$ converges.
Since $0\lt \cos^2 n\lt1$, then $$0\lt \frac{\cos^2 n}{n^2}\lt \frac{1}{n^2}.$$ Now, since $\Sigma_{n=1}^{\infty}\frac{1}{n^2}$ converges, then by the comparison test, the series $\Sigma_{n=1}^{\infty}\frac{\cos^2 n}{n^2}$ converges.