Chapter 10.4, Problem 22E

To determine

To find: Radius of convergence of the given power series.

Answer to Problem 22E

The radius of convergence is $R=\sqrt{2}$

Explanation of Solution

Given:

The given series is,

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(x-\sqrt{2}\right)}^{2n+1}}{2^n}}$

Calculation:

First, using ratio test check the absolute convergence.

So,

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

So,

  $\begin{array}{l}\frac{\left|{\left(x-\sqrt{2}\right)}^2\right|}{2}<1\\ \Rightarrow \left|{\left(x-\sqrt{2}\right)}^2\right|<2\\ \Rightarrow \left|\left(x-\sqrt{2}\right)\right|<\sqrt{2}\end{array}$

Therefore,

If $|x-\sqrt{2}|<\sqrt{2}$ , the given series will converge absolutely and so it is convergent.

Hence, the radius of convergence of the power series given is $R=\sqrt{2}$