# How do you use basic comparison test to determine whether the given series converges or diverges for sum n/sqrt(n^2-1) from n=2 to n=oo?

May 15, 2015

The series diverges.

Let's begin with an inequality :

${n}^{2} - 1 \le {n}^{2}$

$\sqrt{{n}^{2} - 1} \le \sqrt{{n}^{2}} = n$, with $n \ge 1$

$\frac{1}{\sqrt{{n}^{2} - 1}} \ge \frac{1}{n}$, with $n > 1$

$\frac{n}{\sqrt{{n}^{2} - 1}} \ge \frac{n}{n} = 1$

${\sum}_{n = 2}^{\infty} \left(\frac{n}{\sqrt{{n}^{2} - 1}}\right) \ge {\sum}_{n = 2}^{\infty} \left(1\right)$

And obviously, ${\sum}_{n = 2}^{\infty} \left(1\right) = + \infty$, so it diverges.

By comparison test, ${\sum}_{n = 2}^{\infty} \left(\frac{n}{\sqrt{{n}^{2} - 1}}\right)$ diverges too.