Chapter 9, Problem 62RE

To determine

Find sum of the given infinite geometric series

Answer to Problem 62RE

The sum of the infinite geometric series is $\frac{13}{9}$ .

Explanation of Solution

Given:

The geometric sequence,

  ${\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}$

Consider the givengeometric sequence,

  ${\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}$

Write down first few terms

  $1.3+1.3\left(\frac{1}{10}\right)+1.3{\left(\frac{1}{10}\right)}^2+\mathrm{...}$

First find out common ratio r

The common ratio r is can be found out by dividing any term after the first that directly precedes it.

In above geometric sequence,

  $a_1=1.3,a_2=1.3\left(\frac{1}{10}\right)$

Then,

  $\begin{array}{l}r=\frac{a_2}{a_1}\hfill \\ =\frac{1.3\left(\frac{1}{10}\right)}{1.3}\hfill \\ \begin{array}{l} =\frac{1}{10}\\ =0.1\end{array}\hfill \end{array}$

Since $r=0.1$ the condition that $\left|r\right|<1$ is met.

Thus, the infinite geometric series have sum.

The sum S infinite geometric series having common ratio r and first term NA is,

  $S=\frac{a_1}{1-r}\text{ For }\left|r\right|<1$

Then,

  $\begin{array}{l}\begin{array}{l}S=\frac{1.3}{1-0.1}\hfill \\ =\frac{1.3}{0.9}\hfill \end{array} \\ =\frac{13}{9}\end{array}$

Hence, the sum of the infinite geometric series is $\frac{13}{9}$ .