University Calculus: Early Transcendentals (3rd Edition)

We compare with: $\dfrac{1}{n^2+30} \lt \dfrac{1}{n^2}$ We know that $\dfrac{1}{n^2}$ is a convergent p-series with $p=2$ Thus, by the comparison test, $\Sigma_1^\infty \frac{1}{n^2+30}$ converges.