Chapter 8, Problem 16CR

To determine

To find: the sum of infinite series.

Answer to Problem 16CR

Sum of the infinite series ${\displaystyle \sum _{k=1}^{\infty }4{(0.01)}^k}$ is $\underset{\_}{\frac{4}{99}}$ .

Explanation of Solution

Given information:

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

  ${\displaystyle \sum _{k=1}^{\infty }4{(0.01)}^k}$

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.

  ${\displaystyle \sum _{k=1}^{\infty }4{(0.01)}^k}$

Expand the infinite series.

  ${\displaystyle \sum _{k=1}^{\infty }4{(0.01)}^k}=4{(0.01)}^1+4{(0.01)}^2+4{(0.01)}^3+\mathrm{......}$

First term of series is $a_1=4{(0.01)}^1$ .

Second term of series is $a_2=4{(0.01)}^2$ .

Third term of series is $a_3=4{(0.01)}^3$ .

And so on.

Calculating ratio of consecutive terms

  $\frac{a_2}{a_1}=\frac{4{(0.01)}^2}{4{(0.01)}^1}=0.01$ ,

  $\frac{a_3}{a_2}=\frac{4{(0.01)}^3}{4{(0.01)}^2}=0.01$

As ratio $\frac{a_2}{a_1}=\frac{a_3}{a_2}=0.01=\frac{1}{100}$ , so the series is geometric with common ratio that is $r=\frac{1}{100}$ .

Calculating sum of the series

As ${\displaystyle \sum _{k=1}^{\infty }4{(0.01)}^k}=4{(0.01)}^1+4{(0.01)}^2+4{(0.01)}^3+\mathrm{......}$ is infinite geometric series, calculate the sum by using formula of infinite geometric series.

Substitute $a_1=4{(0.01)}^1$ and $r=\frac{1}{100}$ in formula $S_n=\frac{a_1}{1-r}$ .

  $S_n=\frac{4{(0.01)}^1}{1-\left(\frac{1}{100}\right)}$

  $=\frac{\frac{4}{100}}{\left(\frac{100-1}{100}\right)}$

  $=\frac{4}{100}\times \frac{100}{99}$

  $=\frac{4}{99}$

Hence sum of the infinite series ${\displaystyle \sum _{k=1}^{\infty }4{(0.01)}^k}$ is $\underset{\_}{\frac{4}{99}}$ .