Chapter 9, Problem 16RE

a.

To determine

To find: The radius of convergence for the power series ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{x^2-1}{2}\right)}^n}$ .

Answer to Problem 16RE

  $\sqrt{3}$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{x^2-1}{2}\right)}^n}$

Concept Used:

The power series will converge for $\left|x-a\right|<R$ and will diverge for $\left|x-a\right|>R$ .

The radius of convergence is $R$ .

Calculation:

Let us consider $a_n={\left(\frac{x^2-1}{2}\right)}^n$ and $a_{n+1}={\left(\frac{x^2-1}{2}\right)}^{n+1}$ .

Using ratio test find the limit of the ratio of $\left|\frac{a_{n+1}}{a_n}\right|$ as $n\to \infty$ .

  $\begin{array}{l}L=\underset{n\to \infty }{\lim }\left|\frac{{\left(\frac{x^2-1}{2}\right)}^{n+1}}{{\left(\frac{x^2-1}{2}\right)}^n}\right|\left[\text{Simplify}\right]\\ L=\underset{n\to \infty }{\lim }\left(\frac{x^2-1}{2}\right)\\ L=\frac{1}{2}\left|x^2-1\right|\end{array}$

The series is converges if $\frac{1}{2}\left|x^2-1\right|<1$ .

So, the radius of convergence $\left|x^2-1\right|<2\text{or }\left|\left(x-1\right)\left(x+1\right)\right|<2$ .

  $\begin{array}{l}\left|x^2-1\right|<2\\ -2<x^2-1<2\\ -1<x^2<3\\ -1<x^2\left[\text{This is not possible}\right]\\ x^2<3\\ -\sqrt{3}<x<\sqrt{3}\end{array}$

The radius of convergence is $\sqrt{3}$ .

b.

To determine

To find: The interval of convergence for the series ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{x^2-1}{2}\right)}^n}$ .

Answer to Problem 16RE

  $\left(-\sqrt{3},\sqrt{3}\right)$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{x^2-1}{2}\right)}^n}$

Concept Used:

The power series will converge for $\left|x-a\right|<R$ and will diverge for $\left|x-a\right|>R$ .

The radius of convergence is $R$ .

Calculation:

In part (a) calculated the radius of convergence is $R=\sqrt{3}$ .

  $\begin{array}{l}\left|x^2-1\right|<2\\ -2<x^2-1<2\\ -1<x^2<3\\ -1<x^2\left[\text{This is not possible}\right]\\ x^2<3\\ -\sqrt{3}<x<\sqrt{3}\end{array}$

So, the interval of convergence $-\sqrt{3}<x<\sqrt{3}$ .

Now check the end points of the interval.

For $x=-\sqrt{3}$ , the power series ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{x^2-1}{2}\right)}^n}={\displaystyle \sum _{n=1}^{\infty }1}$ is not converges.

For $x=\sqrt{3}$ , the power series ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{x^2-1}{2}\right)}^n}={\displaystyle \sum _{n=1}^{\infty }1}$ is not converges.

Therefore, the interval of convergence is $\left(-\sqrt{3},\sqrt{3}\right)$ .

c.

To determine

To find: The value of $x$ when series ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{x^2-1}{2}\right)}^n}$ converges absolutely.

Answer to Problem 16RE

  $\left(-\sqrt{3},\sqrt{3}\right)$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{x^2-1}{2}\right)}^n}$

Concept Used:

The power series will converge for $\left|x-a\right|<R$ and will diverge for $\left|x-a\right|>R$ .

The radius of convergence is $R$ .

Calculation:

In part (a) calculated the radius of convergence is $R=\sqrt{3}$ .

In part (b) calculated the interval of convergence is $\left(-\sqrt{3},\sqrt{3}\right)$ .

So, the value of $x$ when series converges absolutely is $\left(-\sqrt{3},\sqrt{3}\right)$ .

d.

To determine

To find: The value of $x$ when series ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{x^2-1}{2}\right)}^n}$ converges conditionally.

Answer to Problem 16RE

  $\left(-\sqrt{3},\sqrt{3}\right)$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{x^2-1}{2}\right)}^n}$

Concept Used:

The power series will converge for $\left|x-a\right|<R$ and will diverge for $\left|x-a\right|>R$ .

The radius of convergence is $R$ .

Calculation:

In part (a) calculated the radius of convergence is $R=\sqrt{3}$ .

In part (b) calculated the interval of convergence is $\left(-\sqrt{3},\sqrt{3}\right)$ .

So, the value of $x$ when series converges conditionally is $\left(-\sqrt{3},\sqrt{3}\right)$ .