Chapter 8.5, Problem 20E

To determine

To find: The radius of the convergent and interval of convergence .

Answer to Problem 20E

Radius of convergence $R=3$ and the interval of convergence is $(-\frac{7}{3},\frac{11}{3})$ .

Explanation of Solution

Given information: ${\displaystyle \sum _{n=0}^{\infty }\frac{{(3x-2)}^n}{n{3}^n}}.$and $c_n=\frac{1}{n{3}^n}$

Clearly, the center of the series is

  $3x-2=0$

i.e, $x=\frac{2}{3}$

Now

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{c_n+1}{c_n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{1}{(n+1)·3^{n+1}}·\frac{n·3^n}{1}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{n}{n+1}·\frac{1}{3}\right|\\ =\frac{1}{3}·\underset{n\to \infty }{\lim }\left|\left(\frac{1}{1+\frac{1}{n}}\right)\right|\\ =\frac{1}{3}\end{array}$

By ratio test

Radius of convergence

  $R=\frac{1}{\lim \left|\frac{c_{n+1}}{c_n}\right|}=\frac{1}{\frac{1}{3}}=3$

  $R=3.$

Hence, radius of convergence $R=3.$

And interval of convergence is

  $(-R+c,R+c)$

i.e $(-3+\frac{2}{3},3+\frac{2}{3})$

i.e $(-\frac{7}{3},\frac{11}{3})$

Hence, the interval of convergence is $(-\frac{7}{3},\frac{11}{3})$ .