Past exam question: Convergence of a series or not

Question

1. Is the series $$\sum_{k=1}^{\infty} \frac{\sqrt{k}}{k^2+1}$$ convergent?
   This behaves similar to $\frac{\sqrt{k}}{k^2} = \frac{1}{k^{3/2}}$ so do we use the comparison test???

Answer 1

Note that for all $k\geq 1$, $$ 0\leq\frac{\sqrt{k}}{k^{2}+1}\leq\frac{\sqrt{k}}{k^{2}}=\frac{1}{k^{3/2}}. $$ The series $$ \sum_{k=1}^{\infty}\frac{1}{k^{3/2}} $$ converges since $3/2>1$. Therefore, by the Basic Comparison test, your series also converges.

Answer 2

Yes, since $\frac{\sqrt{k}}{k^2+1} \le \frac{\sqrt{k}}{k^2}$, this is exactly the right approach. Then use the $p$-series test.