Chapter 10.5, Problem 36E

(a)

To determine

To find: The interval of convergence and value of $x$ .

Answer to Problem 36E

  $\left(-1,1\right)$ is the interval of convergence. And value of $x$ lies between $-6<x<-4$ .

Explanation of Solution

Given information: ${\displaystyle \sum _{n=0}^{\infty }{\left(x+5\right)}^n}$

Calculation: The given series is convergent when ${\displaystyle \sum \left|a_n\right|}$ is convergent.

Here, ${\displaystyle \sum a_n}={\displaystyle \sum _{n=0}^{\infty }{\left(x+5\right)}^n}$

  ${\displaystyle \sum \left|a_n\right|}={\displaystyle \sum _{n=0}^{\infty }{\left(x+5\right)}^n}$

  $\begin{array}{l}{a}_n={\left|x+5\right|}^n\\ a_{n+1}={\left|x+5\right|}^{n+1}\\ \frac{a_{n+1}}{a_n}=\frac{{\left|x+5\right|}^{n+1}}{{\left|x+5\right|}^n}\\ \therefore \underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=\underset{n\to \infty }{\lim }\left|x+5\right|\\ =\left|x+5\right|\end{array}$

Therefore, By Ratio test the given series converges for $-6<x<-4$ . And the interval of convergence of $\left(-1,1\right)$ .

(b)

To determine

To find: The interval for absolutely convergence.

Answer to Problem 36E

The series is absolutely convergent for $-6<x<-4$ .

Explanation of Solution

Given information: ${\displaystyle \sum _{n=0}^{\infty }{\left(x+5\right)}^n}$

Calculation: The given series is absolutely convergent when ${\displaystyle \sum \left|a_n\right|}$ is convergent.

Here, ${\displaystyle \sum a_n}={\displaystyle \sum _{n=0}^{\infty }{\left(x+5\right)}^n}$

  ${\displaystyle \sum \left|a_n\right|}={\displaystyle \sum _{n=0}^{\infty }{\left(x+5\right)}^n}$

  $\begin{array}{l}{a}_n={\left|x+5\right|}^n\\ a_{n+1}={\left|x+5\right|}^{n+1}\\ \frac{a_{n+1}}{a_n}=\frac{{\left|x+5\right|}^{n+1}}{{\left|x+5\right|}^n}<1\\ \therefore \left|x+5\right|<1\end{array}$

Therefore, The given series absolutely converges for $-6<x<-4$ .

(c)

To determine

To find: The interval for conditionally convergence.

Answer to Problem 36E

The series is conditionally convergent for no value of $x$ .

Explanation of Solution

Given information: ${\displaystyle \sum _{n=0}^{\infty }{\left(x+5\right)}^n}$

Calculation: The given series is conditionally convergent when ${\displaystyle \sum a_n}$ exists but ${\displaystyle \sum \left|a_n\right|}$ is not.

Here, ${\displaystyle \sum a_n}$ exists for $-6<x<-4$ . And ${\displaystyle \sum \left|a_n\right|}$ also exists for $-6<x<-4$ .

Therefore ${\displaystyle \sum a_n}$ exists so ${\displaystyle \sum \left|a_n\right|}$ exists.

Therefore, the given series diverges conditionally for no value of $x$ , so it converges conditionally for no value of $x$ .