# How to do comparison test for sum 1 / (sqrt(n)) n=1 to n=oo?

May 15, 2015

The series diverges.

Let's begin with an inequality :

$\sqrt{n} \le n ,$, with $n \ge 0$

$\frac{1}{\sqrt{n}} \ge \frac{1}{n}$, with $n \ge 1$

${\sum}_{n = 1}^{\infty} \frac{1}{\sqrt{n}} \ge {\sum}_{n = 1}^{\infty} \frac{1}{n}$

${\sum}_{n = 1}^{\infty} \frac{1}{n}$ is the harmonic series and it diverges.

Therefore, by comparison test, ${\sum}_{n = 1}^{\infty} \frac{1}{\sqrt{n}}$ diverges.