# How do you use the direct Comparison test on the infinite series sum_(n=1)^ooarctan(n)/(n^1.2) ?

Oct 15, 2014

Since

$\arctan n \le \frac{\pi}{2}$,

we have

$\frac{\arctan n}{n} ^ \left\{1.2\right\} \le \frac{\frac{\pi}{2}}{n} ^ \left\{1.2\right\}$.

Since
sum_{n=1}^infty {pi/2}/n^{1.2}=pi/2 sum_{n=1}^infty 1/n^{1.2}

is a convergent p-series with $p = 1.2 > 1$,

by Comparison Test,

${\sum}_{n = 1}^{\infty} \frac{\arctan n}{n} ^ \left\{1.2\right\}$ also converges.

I hope that this was helpful.