Chapter 10, Problem 51RE

To determine

To find:Whether the series ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n\sqrt{n^2-1}}}$ converges absolutely, converges conditionally or diverges.

Answer to Problem 51RE

The series ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n\sqrt{n^2-1}}}$ converges absolutely.

Explanation of Solution

Given:

The given series is, ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n\sqrt{n^2-1}}}$

Using limit comparing test,

On comparing the terms of ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n\sqrt{n^2-1}}}$ with the terms of ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^2}}$ ,

  $\underset{n\to \infty }{\lim }\frac{\frac{1}{n\sqrt{n^2-1}}}{\frac{1}{n^2}}$

  $=\underset{n\to \infty }{\lim }\frac{n^2}{n\sqrt{n^2-1}}$

  $=1$

Now,

For ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^2}}$ using p -test series Test,

  ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^2}}$ converges if $p>1$ , and diverges if $p\le 1$

It is ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^2}}$ , where $p=2$ . Thus, the series converges, as $p>1$

As, the limit is positive and ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^2}}$ converges so, ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n\sqrt{n^2-1}}}$ also converges absolutely.