Use the squeeze theorem to show that lim as n goes to infinity arctan(n^2)/Sqrt(n) = 0
note that $\displaystyle - \frac {\pi}2 \le \arctan x \le \frac {\pi}2$ for all $\displaystyle x$, thus we have
$\displaystyle - \frac {\pi}{2 \sqrt{n}} \le \frac {\arctan (n^2)}{\sqrt{n}} \le \frac {\pi}{2 \sqrt{n}}$
can you finish?
(also, note that we can make the left side zero, but this looks nice for symmetry )