Chapter 8.6, Problem 2E

To determine

To explain: The behavior of the series ${\displaystyle \sum _{n=0}^{\infty }\frac{b_n}{n+1}{x}^{n+1}}$ .

Answer to Problem 2E

The series ${\displaystyle \sum _{n=0}^{\infty }\frac{b_n}{n+1}{x}^{n+1}}$ is convergent and the interval of convergence is $\left|x\right|<2$ .

Explanation of Solution

Given information: The series ${\displaystyle \sum _{n=0}^{\infty }{b}_n{x}^n}$ converges for $\left|x\right|<2$ .

Calculation:

Comparing ${\displaystyle \sum _{n=0}^{\infty }{b}_n{x}^n}$ with ${\displaystyle \sum _{n=0}^{\infty }{a}_n}$ ,

  $a_n=b_n{x}^n,a_{n+1}=b_{n+1}{x}^{n+1}$ .

  $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{b_{n+1}{x}^{n+1}}{b_n{x}^n}\right|=\underset{n\to \infty }{\lim }\left|\frac{b_{n+1}}{b_n}x\right|$

By Ratio Test, the series ${\displaystyle \sum _{n=0}^{\infty }{a}_n}$ converges if $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|<1$ .

So, $\underset{n\to \infty }{\lim }\left|\frac{b_{n+1}}{b_n}x\right|<1$ . −(i)

Since the interval of convergence is $\left|x\right|<2$ , so $\underset{n\to \infty }{\lim }\frac{b_{n+1}}{b_n}=\frac{1}{2}$ .

Now, for the series ${\displaystyle \sum _{n=0}^{\infty }\frac{b_n}{n+1}{x}^{n+1}}$ ,

  $a_n=\frac{b_n}{n+1}{x}^{n+1},a_{n+1}=\frac{b_{n+1}}{n+2}{x}^{n+2}$ .

  $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{\frac{b_{n+1}}{n+2}{x}^{n+2}}{\frac{b_n}{n+1}{x}^{n+1}}\right|=\underset{n\to \infty }{\lim }\left|\frac{n+1}{n+2}\frac{b_{n+1}}{b_n}x\right|<1$ (Using (i))

  $\begin{array}{l}\Rightarrow \underset{n\to \infty }{\lim }\left|\frac{1+\frac{1}{n}}{1+\frac{2}{n}}\frac{b_{n+1}}{b_n}x\right|<1\\ \Rightarrow \left|\frac{1+0}{1+0}\frac{1}{2}x\right|<1\\ \Rightarrow \left|\frac{x}{2}\right|<1\\ \Rightarrow \left|x\right|<2\end{array}$

Thus, the series ${\displaystyle \sum _{n=0}^{\infty }\frac{b_n}{n+1}{x}^{n+1}}$ is convergent and the interval of convergence is $\left|x\right|<2$ .