I don't agree with your upper bound...
$\displaystyle -\frac{\pi}{2} < \arctan{n} < \frac{\pi}{2}$ for all $\displaystyle n$, so surely
$\displaystyle -\frac{\frac{\pi}{2}}{n\sqrt{n}} < \frac{\arctan{n}}{n\sqrt{n}} < \frac{\frac{\pi}{2}}{n\sqrt{n}}$ (we can assume $\displaystyle n\sqrt{n} > 0$ since we are summing from $\displaystyle n = 1$ to $\displaystyle \infty$...
So you will need to choose $\displaystyle \frac{\frac{\pi}{2}}{n\sqrt{n}}$ as your upper bound.
The rest of your logic looks fine.