Chapter 9.5, Problem 4QQ

(a)

To determine

To find: All the values of $x$ for which the given series converges absolutely.

Answer to Problem 4QQ

The values of $x$ for which the given series converges absolutely is $-2<x<-1$ that is , on $\left(-2,-1\right)$ .

Explanation of Solution

Given information: The series ${\displaystyle \sum _{n=0}^{\infty }\frac{n{\left(2x+3\right)}^n}{n+2}}$ is given.

Concept used :

If the series ${\displaystyle \sum a_n}$ is with non-negative terms and with $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=L$ , then

• the series is said to be convergent if $L<1$ .
• the series is said to be divergent if $L>1$ .
• The test is said to be inconclusive if $L=1$

According to $n^{\text{th}}$ -Term Test if $\underset{n\to \infty }{\lim }{a}_n\ne 0$ or fails to exist, then ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ diverges.

Calculation:

Determine the absolute convergence using the ratio test for the give series where $a_n=\frac{n{\left(2x+3\right)}^n}{n+2}$ .

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{\frac{\left(n+1\right)\left|{\left(2x+3\right)}^{n+1}\right|}{n+3}}{\frac{n\left|{\left(2x+3\right)}^n\right|}{n+2}}\\ =\underset{n\to \infty }{\lim }\frac{\left(n+1\right)\left|{\left(2x+3\right)}^{n+1}\right|}{n+3}\cdot \frac{n+2}{n\left|{\left(2x+3\right)}^n\right|}\\ =\underset{n\to \infty }{\lim }\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+3\right)}\left|\left(2x+3\right)\right|\\ =\underset{n\to \infty }{\lim }\left(\frac{n^2+3n+2}{n^2+3n}\right)\left|\left(2x+3\right)\right|\\ =\underset{n\to \infty }{\lim }\left(\frac{1+\frac{3}{n}+\frac{2}{n^2}}{1+\frac{3}{n}}\right)\left|\left(2x+3\right)\right|\\ =\frac{\underset{n\to \infty }{\lim }\left(1+\frac{3}{n}+\frac{2}{n^2}\right)}{\underset{n\to \infty }{\lim }\left(1+\frac{3}{n}\right)}\left|\left(2x+3\right)\right|\\ =\frac{1}{1}\left|\left(2x+3\right)\right|\\ =\left|\left(2x+3\right)\right|\end{array}$

Since the series is convergent if $L<1$ , $\left|2x+3\right|<1$ .

Simplify the inequality to obtain the required result by applying the absolute rule, if $\left|u\right|<a$ , $a>0$ then $-a<u<a$ to $\left|2x+3\right|<1$ .

  $\begin{array}{c}-1<2x+3<1\\ 2x+3>-1\text{ and 2}x+3<1\\ x>-2\text{and}x<-1\end{array}$

Merge overlapping intervals.

  $-2<x<-1$

Thus, the series converges absolutely for $-2<x<-1$ , that is on the interval $\left(-2,-1\right)$ .

(b)

To determine

To find: All the values of $x$ for which the given series converges conditionally.

Answer to Problem 4QQ

There are no values of $x$ for which the given series converges conditionally.

Explanation of Solution

Given information: The series ${\displaystyle \sum _{n=0}^{\infty }\frac{n{\left(2x+3\right)}^n}{n+2}}$ is given.

Concept used :

According to $n^{\text{th}}$ -Term Test if $\underset{n\to \infty }{\lim }{a}_n\ne 0$ or fails to exist, then ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ diverges.

Calculation:

Check the convergence at the end points.

Determine $\underset{n\to \infty }{\lim }\frac{n{\left(2x+3\right)}^n}{n+2}$ at $x=-2$ by substituting $x=-2$ into $\underset{n\to \infty }{\lim }\frac{n{\left(2x+3\right)}^n}{n+2}$ .

  $\begin{array}{c}{\left.\underset{n\to \infty }{\lim }\frac{n{\left(2x+3\right)}^n}{n+2}\right|}_{x=-2}=\underset{n\to \infty }{\lim }\frac{n{\left(2\left(-2\right)+3\right)}^n}{n+2}\\ =\underset{n\to \infty }{\lim }\frac{n{\left(-4+3\right)}^n}{n+2}\\ =\underset{n\to \infty }{\lim }\frac{n{\left(-1\right)}^n}{n+2}\\ =\underset{n\to \infty }{\lim }\frac{{\left(-1\right)}^n}{1+\frac{2}{n}}\\ =\text{DNE}\\ \ne 0\end{array}$

Since $\underset{n\to \infty }{\lim }\frac{n{\left(2x+3\right)}^n}{n+2}\ne 0$ , ${\displaystyle \sum _{n=0}^{\infty }\frac{n{\left(2x+3\right)}^n}{n+2}}$ diverges at $x=-2$ .

Determine $\underset{n\to \infty }{\lim }\frac{n{\left(2x+3\right)}^n}{n+2}$ at $x=-1$ by substituting $x=-1$ into $\underset{n\to \infty }{\lim }\frac{n{\left(2x+3\right)}^n}{n+2}$ .

  $\begin{array}{c}{\left.\underset{n\to \infty }{\lim }\frac{n{\left(2x+3\right)}^n}{n+2}\right|}_{x=-2}=\underset{n\to \infty }{\lim }\frac{n{\left(2\left(-1\right)+3\right)}^n}{n+2}\\ =\underset{n\to \infty }{\lim }\frac{n{\left(-2+3\right)}^n}{n+2}\\ =\underset{n\to \infty }{\lim }\frac{n{\left(1\right)}^n}{n+2}\\ =\underset{n\to \infty }{\lim }\frac{{\left(1\right)}^n}{1+\frac{2}{n}}\\ =1\\ \ne 0\end{array}$

Since $\underset{n\to \infty }{\lim }\frac{n{\left(2x+3\right)}^n}{n+2}\ne 0$ , ${\displaystyle \sum _{n=0}^{\infty }\frac{n{\left(2x+3\right)}^n}{n+2}}$ diverges at $x=-1$ .

It is clearly observable that the given series converges absolutely on $\left(-2,-1\right)$ and diverges at both the end points. Therefore, there is no values of $x$ for which the given series converges conditionally.