Chapter 10.1, Problem 24E

To determine

To calculate: The interval of convergence and write the function in terms of $x$ .

Answer to Problem 24E

The required function is $f\left(x\right)=\frac{6}{3+x}$ .

Explanation of Solution

Given Information:

The series is ${\displaystyle \sum _{n=0}^{\infty }3{\left(\frac{x-1}{2}\right)}^n}$ .

Formula Used:

The function of the geometric series is $f\left(x\right)=\frac{a}{1-r}$ , where $a$ is initial term and $r$ is the common ratio.

Interval of Convergence:

The geometric series converges then, the interval of convergence is $\left|r\right|<1$ .

Calculation:

The series is ${\displaystyle \sum _{n=0}^{\infty }3{\left(\frac{x-1}{2}\right)}^n}$ .

In the series ${\displaystyle \sum _{n=0}^{\infty }3{\left(\frac{x-1}{2}\right)}^n}$ , $a=1$ and $r=\left(\frac{x-1}{2}\right)$ .

Now, find the interval of convergence.

  $\begin{array}{c}\left|\frac{x-1}{2}\right|<1\\ -1<\frac{x-1}{2}<1\\ -2<x-1<2\\ -2+1<x-1+1<2+1\\ -1<x<3\end{array}$

Find the function in terms of $x$ .

Substitute 1 for $a$ and $\left(\frac{x-1}{2}\right)$ for $r$ in the formula $\frac{a}{1-r}$ .

  $\begin{array}{c}f\left(x\right)=3\frac{1}{1-\left(\frac{x-1}{2}\right)}\\ =\frac{3}{\left(\frac{2-x+1}{2}\right)}\\ =\frac{3}{\left(\frac{3+x}{2}\right)}\\ =\frac{6}{3+x}\end{array}$ , where $-1<x<3$

Hence, the requiredfunction is $f\left(x\right)=\frac{6}{3+x}$ .