## University Calculus: Early Transcendentals (3rd Edition)

Consider $a_n=\dfrac{1+\cos n}{n^2}$ We know that $-1 \leq \cos^2 n \leq 1$ Now, $a_n \leq \dfrac{2}{n^2}$ and $\Sigma_{n=1}^\infty \dfrac{2}{n^2}$ is a convergent p-series with $p=2$ Hence, the series converges due to the direct comparison test.