Answer
Converges
Work Step by Step
Here, $\Sigma_{n=1}^\infty \dfrac{\sqrt n}{n^2+1} \lt \Sigma_{n=1}^\infty\dfrac{\sqrt n}{n^2}=\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}}$
But the series $\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}}$ shows a convergent $p$-series test with $p=\dfrac{3}{2} \gt 1$
Hence, the series converges by the direct comparison test.