Chapter 10.5, Problem 42E

To determine

The internal of the convergence of the series ${\displaystyle \sum _{n=0}^{\infty }\frac{x^{2n+1}}{n!}}$

Answer to Problem 42E

The Interval of convergence is $\left[-\infty ,\infty \right]$

Explanation of Solution

Given information:

The given series is

  ${\displaystyle \sum _{n=0}^{\infty }\frac{x^{2n+1}}{n!}}$

Formula used:

  ${\displaystyle \sum _{n=0}^{\infty }\left|a_n\right|}={\displaystyle \sum _{n=0}^{\infty }\frac{{\left|x\right|}^{2n+1}}{n!}}$

Calculation:

The series of absolute values is

  ${\displaystyle \sum _{n=0}^{\infty }\left|a_n\right|}={\displaystyle \sum _{n=0}^{\infty }\frac{{\left|x\right|}^{2n+1}}{n!}}$

The ratio Test let ${\displaystyle \sum a_n}$ be a series with positive terms, and with

  $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=L$

Then, the series converges if L < I, the series diverges if L >1 and the test is inconclusive if L=1

Using the ratio test, we check for absolute convergence as follows,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{{\left|x\right|}^{2n+3}}{\left(n+1\right)!}\times \frac{n!}{{\left|x\right|}^{2n+1}}\\ =x^2\underset{n\to \infty }{\lim }\frac{1}{n+1}\\ \underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=0\end{array}$

The series converges absolutely for all real numbers.

(a) Interval of convergence: $\left[-\infty ,\infty \right]$

(b) Series converges absolutely on $\left[-\infty ,\infty \right]$

(c) None

Conclusion:

The Interval of convergence is $\left[-\infty ,\infty \right]$