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

Since, we have $\Sigma_{n=1}^\infty \dfrac{\sqrt n}{n^2+1} \lt \Sigma_{n=1}^\infty\dfrac{\sqrt n}{n^2}$ or, $=\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}}$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}}$ is a convergent series due to the p-series test with $p=\dfrac{3}{2} \gt 1$ Thus the series converges by the direct comparison test.