Chapter 10, Problem 1RE

(a)

To determine

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

Answer to Problem 1RE

The series is convergence for all real x. Therefore, the radius of convergence for is $\infty$ . $R=\infty$

Explanation of Solution

Given information:

The given function is,

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

Formula used:

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{\left|{\left(-x\right)}^{n+1}\right|}{\left(n+1\right)!}\frac{n!}{\left|{\left(-x\right)}^n\right|}$

Calculation:

Check for absolute convergence using ratio test. $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{\left|{\left(-x\right)}^{n+1}\right|}{\left(n+1\right)!}\frac{n!}{\left|{\left(-x\right)}^n\right|}$

  $\begin{array}{l}=\underset{n\to \infty }{\lim }\frac{1}{n+1},\left|-x\right|\\ =\left|-x\right|\underset{n\to \infty }{\lim }\frac{1}{n+1}\\ =\left|-x\right|.0\\ =0\end{array}$

Therefore, the series is convergence for all real x. Therefore, the radius of convergence for is $\infty$ . $R=\infty$

Conclusion:

The series is convergence for all real x. Therefore, the radius of convergence for is $\infty$ . $R=\infty$

(b)

To determine

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

Answer to Problem 1RE

The series always converges for interval $\left(-\infty ,\infty \right)$

Explanation of Solution

Given information:

The given function is,

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

Formula used:

Ratio test is used.

Calculation:

Since,

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=0$

Therefore the series always absolutely for all real x

Since,

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}<1$

Therefore, the series always converges for interval $\left(-\infty ,\infty \right)$

Conclusion:

The series always converges for interval $\left(-\infty ,\infty \right)$

(c)

To determine

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

Answer to Problem 1RE

The series converges absolutely for all x.

Explanation of Solution

Given information:

The given function is,

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

Formula used:

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=0$

Calculation:

Since,

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=0$

Therefore the series always converges absolutely for all real x

Since,

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}<1$

Therefore, the series is always converges absolutely for interval $\left(-\infty ,\infty \right)$

Conclusion:

The series converges absolutely for all x.

(d)

To determine

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

Answer to Problem 1RE

The series always converges absolutely for all real x, there is no point of conditional convergence.

Explanation of Solution

Given information:

The given function is,

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

Formula used:

Ratio test is used.

Calculation:

Since, the series always converges absolutely for all real x, there is no point of conditional convergence.

Conclusion:

The series always converges absolutely for all real x, there is no point of conditional convergence.