Chapter 8, Problem 64RE

To determine

To calculate: To find the sum of the infinite geometric series, if possible and if not possible, explain why

Answer to Problem 64RE

  ${\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}=\frac{13}{9}$

Explanation of Solution

Given information: Geometric series is ${\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}$

Formula Used:

Sum of infinite geometric series is given as

  $S_{\infty }=\frac{a}{1-r}$

Where,

  $a=$ First term

  $r=$ Common ratio

Calculation:

Geometric series is given as follows:

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

On simplifying,

  $\begin{array}{l}{\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}={\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^k{\left(\frac{1}{10}\right)}^{-1}}\\ {\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}={\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^k\cdot \frac{10}{1}}\\ {\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}=1.3\times 10{\displaystyle \sum _{k=1}^{\infty }{\left(\frac{1}{10}\right)}^k}\\ {\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}=13{\displaystyle \sum _{k=1}^{\infty }{\left(\frac{1}{10}\right)}^k}\\ {\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}=13\left[\frac{1}{10}+{\left(\frac{1}{10}\right)}^2+{\left(\frac{1}{10}\right)}^3+{\left(\frac{1}{10}\right)}^4+\mathrm{.....}\right]\end{array}$

Here,

  $\begin{array}{l}a=\frac{1}{10}\\ r=\frac{1}{10}\end{array}$

Sum of infinite geometric series is given as

  $S_{\infty }=\frac{a}{1-r}$

Thus, sum is given as

  $\begin{array}{l}{\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}=13\left[\frac{\frac{1}{10}}{1-\frac{1}{10}}\right]\\ {\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}=13\left[\frac{\frac{1}{10}}{\frac{10-1}{10}}\right]\\ {\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}=13\cdot \frac{1}{9}\\ {\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}=\frac{13}{9}\end{array}$

Conclusion:

Hence, ${\displaystyle \sum _{k=1}^{\infty }1.3{\left(\frac{1}{10}\right)}^{k-1}}=\frac{13}{9}$