Answer
Converges
Work Step by Step
We will have to compare with $\dfrac{1}{n^2+30} \lt \dfrac{1}{n^2}$
$\dfrac{1}{n^2}$ shows a convergent p-series with p= 2.
Hence, by the comparison test, we have $\Sigma_1^{\infty}\dfrac{1}{n^2+30} $ is a convergent series