Chapter 9.3, Problem 71E

To determine

To find the sum of the infinite geometric series ${\displaystyle {\sum }_{n=0}^{\infty }{\left(-\frac{1}{2}\right)}^n}$ .

Answer to Problem 71E

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

Explanation of Solution

Given information :

Let the infinite geometric series be defined as ${\displaystyle {\sum }_{n=0}^{\infty }{\left(-\frac{1}{2}\right)}^n}$ .

Concept used:

In a sequence $a_1,a_2,a_3,a_4,\dots \dots ,a_n,\dots$ if ratio of the consecutive terms is the same then it is defined as Geometric Sequence.

So, a number r is defined as the common ratio of the geometric sequence if $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\dots \dots =r.$

The sum of an infinite geometric series $a_1+a_{1}r+a_1{r}^2+a_1{r}^3+\cdots +a_1{r}^{n-1}+\cdots$ is $S={\displaystyle {\sum }_{i=0}^{\infty }{a}_1{r}^i=\frac{a_1}{1-r}}$ where $\left|r\right|<1$ .

Calculations:

  $\begin{array}{c}{\displaystyle {\sum }_{n=0}^{\infty }{\left(-\frac{1}{2}\right)}^n}={\left(-\frac{1}{2}\right)}^0+{\left(-\frac{1}{2}\right)}^1+{\left(-\frac{1}{2}\right)}^2+\cdots +{\left(-\frac{1}{2}\right)}^n+\cdots \\ =1+\left(-\frac{1}{2}\right)+{\left(-\frac{1}{2}\right)}^2+\cdots +{\left(-\frac{1}{2}\right)}^n+\cdots \\ =1-\frac{1}{2}+\frac{1}{4}+\cdots +{\left(-\frac{1}{2}\right)}^n+\cdots \end{array}$

Here, $a_1=1$ and common ratio $r=\frac{a_2}{a_1}=\frac{-\frac{1}{2}}{1}=-\frac{1}{2}$ . In order to find out the sum of the series ${\displaystyle {\sum }_{n=0}^{\infty }{\left(-\frac{1}{2}\right)}^n}$ , substitute these values in $\frac{a_1}{1-r}$ and get,

  $\begin{array}{c}S=\frac{1}{1-\left(-\frac{1}{2}\right)}\\ =\frac{1}{\left(\frac{3}{2}\right)}\\ =\frac{2}{3}\end{array}$

Hence, ${\displaystyle {\sum }_{n=0}^{\infty }{\left(-\frac{1}{2}\right)}^n}=\frac{2}{3}$ .