Chapter 10.4, Problem 18E

To determine

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

Answer to Problem 18E

The radius of convergence is $R=3$

Explanation of Solution

Given:

The given series is,

  ${\displaystyle \sum _{n=0}^{\infty }\frac{\sqrt{n}{x}^n}{3^n}}$

Calculation:

First, using ratio test check the absolute convergence.

So,

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

So,

  $\begin{array}{l}\frac{\left|x\right|}{3}<1\\ \Rightarrow \left|x\right|<3\end{array}$

Therefore,

If $\left|x\right|<3$ , the given series will converge absolutely and so it is convergent.

Hence, the radius of convergence of the given power series is $R=3$