Chapter 8, Problem 9RQ

To determine

To find: The statement is true or false and if the statement then given the explanation.

Answer to Problem 9RQ

The given statement is true

Explanation of Solution

Given:

The given statement is that if $0\le a_n\le b_n$ and $b_n$ diverges, then ${\displaystyle \sum a_n}$ diverges as $a_n=\frac{n}{n^2+1}$ and $b_n=\frac{1}{n^2}$ , then $a_n\le b_n$ for all values n .

Calculation:

Consider that the series is ${\displaystyle \sum _{n=1}^{\infty }\frac{n}{n^2+1}}$ .

Then, by the comparison test,

  $\begin{array}{l}\frac{1}{n^2+1}\le \frac{1}{n^2}\\ \frac{n}{n^2+1}\le \frac{n}{n^3}=\frac{1}{n^2}\end{array}$

For, $a_n=\frac{n}{n^2+1}$ and $b_n=\frac{1}{n^2}$ , then $a_n\le b_n$ for all values n .

Then, ${\displaystyle \sum _{n=0}^{\infty }{b}_n}={\displaystyle \sum _{n=0}^{\infty }\frac{1}{n^2}}$ is the series with is a p series with $p=2>1$ .

The,

  ${\displaystyle \sum _{n=0}^{\infty }{b}_n}$

The above expression is convergent.

Then by the comparison test series ${\displaystyle \sum _{n=0}^{\infty }\frac{n}{n^2+1}}$ convergent.