## Multivariable Calculus, 7th Edition

$a_{n}=arctan(n)$ $\lim\limits_{n \to \infty}a_{n} = \lim\limits_{n \to \infty} arctan(n)$ $=\frac{\pi}{2}$ 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.