Answer
The series is divergent.
Work Step by Step
$a_{n}=arctan(n)$
$\lim\limits_{n \to \infty}a_{n} = \lim\limits_{n \to \infty} arctan(n)$
$=\frac{\pi}{2}$
since [$\lim\limits_{x \to \infty} (tan)^{-1}x = \frac{\pi}{2}$]
$\lim\limits_{n \to \infty}a_{n}=\frac{\pi}{2} \ne 0$
So the series is divergent.