Chapter 8, Problem 15CR

To determine

To find: the sum of infinite series.

Answer to Problem 15CR

Sum of the infinite series ${\displaystyle \sum _{j=1}^{\infty }\frac{8}{{10}^j}}$ is $\underset{\_}{\frac{8}{9}}$ .

Explanation of Solution

Given information:

The infinite series whose sum needs to be calculated is as follows

  ${\displaystyle \sum _{j=1}^{\infty }\frac{8}{{10}^j}}$

Formula used:

Sum of infinite geometric series $S_n=\frac{a_1}{1-r}$ where $a_1$ is first term and $r$ is common ratio.

Calculation:

The infinite series is as follows.

  ${\displaystyle \sum _{j=1}^{\infty }\frac{8}{{10}^j}}$

Expand the infinite series.

  ${\displaystyle \sum _{j=1}^{\infty }\frac{8}{{10}^j}}=\frac{8}{10}+\frac{8}{{10}^2}+\frac{8}{{10}^3}+\mathrm{......}$

First term of series is $a_1=\frac{8}{10}$ .

Second term of series is $a_2=\frac{8}{{10}^2}$ .

Third term of series is $a_3=\frac{8}{{10}^3}$ .

And so on.

Calculating ratio of consecutive terms

  $\frac{a_2}{a_1}=\frac{\left(\frac{8}{{10}^2}\right)}{\left(\frac{8}{10}\right)}=\frac{1}{10}$ ,

  $\frac{a_3}{a_2}=\frac{\left(\frac{8}{{10}^3}\right)}{\left(\frac{8}{{10}^2}\right)}=\frac{1}{10}$

As ratio $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{1}{10}$ , so the series is geometric with common ratio that is $r=\frac{1}{10}$ .

Calculating sum of the series

As ${\displaystyle \sum _{j=1}^{\infty }\frac{8}{{10}^j}}=\frac{8}{10}+\frac{8}{{10}^2}+\frac{8}{{10}^3}+\mathrm{......}$ is infinite geometric series, calculate the sum by using formula of infinite geometric series.

Substitute $a_1=\frac{8}{10}$ and $r=\frac{1}{10}$ in formula $S_n=\frac{a_1}{1-r}$ .

  $S_n=\frac{\left(\frac{8}{10}\right)}{1-\left(\frac{1}{10}\right)}$

  $=\frac{\left(\frac{8}{10}\right)}{\left(\frac{10-1}{10}\right)}$

  $=\frac{8}{10}\times \frac{10}{9}$

  $=\frac{8}{9}$

Hence sum of the infinite series ${\displaystyle \sum _{j=1}^{\infty }\frac{8}{{10}^j}}$ is $\underset{\_}{\frac{8}{9}}$ .