## Calculus 8th Edition

nth term test of divergence: if $\lim\limits_{n \to \infty} {a_{n}} \ne 0$, then the series $\Sigma a_{n}$ diverges. In the problem $a_{n} = arctan(n)$ $\lim\limits_{n \to \infty} arctan(n) = \frac{\pi}{2}$ Since $\frac{\pi}{2} \ne 0$, the series $\sum_{n}^{\infty} \arctan n$ diverges.