Chapter 9, Problem 29CLT

To determine

To find:The sum of the series ${\displaystyle \sum _{i=0}^{\infty }1.3{\left(\frac{1}{10}\right)}^{i-1}}$ .

Answer to Problem 29CLT

The sum is $\frac{130}{9}$ .

Explanation of Solution

Let $a_n={\displaystyle \sum _{i=0}^{\infty }1.3{\left(\frac{1}{10}\right)}^{i-1}}$ .

In order to find the sum of the series ${\displaystyle \sum _{i=0}^{\infty }1.3{\left(\frac{1}{10}\right)}^{i-1}}$ , first check the first few terms of the series to determine the type of the series.

  $\begin{array}{c}{a}_1=1.3{\left(\frac{1}{10}\right)}^{0-1}\\ =1.3{\left(\frac{1}{10}\right)}^{-1}\\ =13\end{array}$

  $\begin{array}{c}{a}_2=1.3{\left(\frac{1}{10}\right)}^{1-1}\\ =1.3{\left(\frac{1}{10}\right)}^0\end{array}$

  $\begin{array}{c}{a}_3=1.3{\left(\frac{1}{10}\right)}^{2-1}\\ =1.3{\left(\frac{1}{10}\right)}^1\end{array}$

Observe that the given series is a geometric series with first term $a_1=1.3$ and common ratio $r=\frac{1}{10}$ .

Since $\left|r\right|<1$ , the formula for sum to infinity of a geometric series can be applied here.

So, the sum of the series is:

  $\begin{array}{c}{\displaystyle \sum _{i=0}^{\infty }1.3{\left(\frac{1}{10}\right)}^{i-1}}=\frac{a_1}{1-r}\\ =\frac{13}{1-\frac{1}{10}}\\ =\frac{13\times 10}{9}\\ =\frac{130}{9}\end{array}$