Chapter 9, Problem 1RE

a.

To determine

To find: The radius of convergence for the series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-x\right)}^n}{n!}}$ .

Answer to Problem 1RE

  $\infty$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-x\right)}^n}{n!}}$

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

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

  $\begin{array}{l}L=\underset{n\to \infty }{\lim }\left|\frac{\frac{{\left(-x\right)}^{n+1}}{\left(n+1\right)!}}{\frac{{\left(-x\right)}^n}{n!}}\right|\left[\text{Simplify}\right]\\ L=\underset{n\to \infty }{\lim }\left|\frac{\left(-x\right)}{n}\right|\\ L=\left|x\right|\underset{n\to \infty }{\lim }\frac{1}{n}\\ L=0\end{array}$

Since $L<1$ for all the values of $x$ , the series will converge for all real value of $x$ .

The radius of convergence is $\infty$ .

b.

To determine

To find: The interval of convergence for the series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-x\right)}^n}{n!}}$ .

Answer to Problem 1RE

  $\left(-\infty ,\infty \right)$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-x\right)}^n}{n!}}$

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 $R=\infty$ .

So, the interval of convergence would be $\left(-\infty ,\infty \right)$ .

c.

To determine

To find: The value of $x$ when series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-x\right)}^n}{n!}}$ converges absolutely.

Answer to Problem 1RE

  $-\infty <x<\infty$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-x\right)}^n}{n!}}$

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 $R=\infty$ .

In part (b) calculated the interval of convergence is $\left(-\infty ,\infty \right)$ .

So, the value of $x$ when series converges absolutely is $-\infty <x<\infty$ .

d.

To determine

To find: The value of $x$ when series ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-x\right)}^n}{n!}}$ converges conditionally.

Answer to Problem 1RE

  $-\infty <x<\infty$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-x\right)}^n}{n!}}$

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 $R=\infty$ .

In part (b) calculated the interval of convergence is $\left(-\infty ,\infty \right)$ .

So, the value of $x$ when series converges conditionally is $-\infty <x<\infty$ .