Chapter 8.4, Problem 24E

To determine

To find: The series is absolute convergent or divergent.

Answer to Problem 24E

  ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{\sqrt[n]{n}}{n+1}}$is not absolute convergent.

Explanation of Solution

Given information: ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{\sqrt[n]{n}}{n+1}}$

At first we take absolute value of ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{\sqrt[n]{n}}{n+1}}$

  $\text{is }{\displaystyle \sum _{n=1}^{\infty }\frac{\sqrt{n}}{n+1}}$

Now, let, $a_n=\frac{\sqrt{n}}{n+1}$

We take the limit comparison test $b_n=\frac{\sqrt{n}}{n}=\frac{1}{\sqrt{n}}$

Now, $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\underset{n\to \infty }{\lim }\frac{\sqrt{n}}{n+1}\cdot \sqrt{n}\\ \text{ =1 (positive real number)}\end{array}$

Then by the limit comparison test $a_n{\text{ and b}}_n$ both converge or diverge together .

Since, $b_n=\frac{1}{\sqrt{n}}$ is divergent ( by p- series with $p=\frac{1}{2}<1)$

Then by limit comparison test ${\displaystyle \sum a_n}$ is divergent.

Hence, ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{\sqrt[n]{n}}{n+1}}$ is not absolute convergent.