Determine whether $\sum\limits_{n=1}^{\infty} \frac{ \sqrt{n} }{n^2+3}$ converges or diverges. - Please check my reasoning. Determine whether $\sum_{n=1}^{\infty} \dfrac{ \sqrt{n} }{n^2+3}$ converges or diverges.
I would appreciate it if someone could check my reasoning for this problem and indicate if I have made any mistakes, why they are mistakes, and how to properly reason for the answer.
My reasoning is as follows.
We know that $ 0 > \dfrac{ \sqrt{n} }{n^2+3} > \dfrac{1}{n^2} $.
We also know that $\dfrac{1}{n^2}$ diverges by the p-series test, since $p=2>1$.
Therefore, since $\dfrac{1}{n^2}$ diverges, $\sum_{n=1}^{\infty} \dfrac{ \sqrt{n} }{n^2+3}$ also must diverge, since $0 >\dfrac{ \sqrt{n} }{n^2+3} > \dfrac{1}{n^2}$.
Thank you.

One may observe that, for $n>1$,
$$
0<\frac{\sqrt{n}}{n^2+3} < \frac{\sqrt{n}}{n^2}=\frac{1}{n^{3/2}}
$$ since  $\displaystyle \sum\frac{1}{n^{3/2}}$ is convergent then the initial series $\displaystyle \sum\frac{\sqrt{n}}{n^2+3} $ is convergent by the comparison test.
