Chapter 8.5, Problem 34E

To determine

To calculate: The radius of convergence of the power series.

Answer to Problem 34E

The radius of convergence of the power series is $\sqrt{R}$ .

Explanation of Solution

Given information:

The radius of convergence of the series ${\displaystyle \sum c_n{x}^n}$ is $R$ .

Concept Used:

Ratio Test:

(i) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L<1$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$ is absolutely convergent.

(and therefore convergent)

(ii) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L>1$ or $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\infty$ , then the series ${\displaystyle \sum _{n=1}^{\infty }{a}_n}$is divergent.

(iii) If $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=1$, the Ratio test is inconclusive.

Calculation:

The series given is

  ${\displaystyle \sum c_n{x}^n}$

Then apply ratio test to this

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=L\\ L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}{x}^{n+1}}{c_n{x}^n}\right|\\ =\left|x\right|\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}}{c_n}\right|\end{array}$

The series is conversent if the value of $L$ is $L<1$

Therefore

  $\begin{array}{l}\left|x\right|\cdot \underset{n\to \infty }{\lim }\left|\frac{c_{n+1}}{c_n}\right|<1\\ \left|x\right|<\frac{1}{\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}}{c_n}\right|}\end{array}$

So, the radius of convergence $R$ is

  $R=\frac{1}{\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}}{c_n}\right|}$

Then

  $\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}}{c_n}\right|=\frac{1}{R}$

For the series ${\displaystyle \sum c_n{x}^n}$ apply ratio test

  $\begin{array}{l}L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}{x}^{2\left(n+1\right)}}{c_n{x}^{2n}}\right|\\ =x^2\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}}{c_n}\right|\\ \end{array}$

Put the value in this equation

  $\underset{n\to \infty }{\lim }\left|\frac{c_{n+1}}{c_n}\right|=\frac{1}{R}$

Then,

  $\begin{array}{l}L=x^2\cdot \frac{1}{R}\\ \end{array}$

The series is conversent if the value of $L$ is $L<1$

Therefore,

  $\begin{array}{l}\frac{x^2}{R}<1\\ x^2<R\\ \text{-}\sqrt{\text{R}}<x<\sqrt{\text{R}}\\ \end{array}$

So, the radius of convergence of the power series is $\sqrt{R}$

Conclusion:

The radius of convergence of the power series is $\sqrt{R}$