Chapter 7.4, Problem 20E

Solution

To calculate : The sum of infinite geometric series $-\frac{1}{8}-\frac{1}{12}-\frac{1}{18}-\frac{1}{27}+\cdots$ if it exists.

The sum of geometric series $-\frac{1}{8}-\frac{1}{12}-\frac{1}{18}-\frac{1}{27}+\cdots$ is $-\frac{3}{8}$ .

Given information :

The geometric series $-\frac{1}{8}-\frac{1}{12}-\frac{1}{18}-\frac{1}{27}+\cdots$ .

Formula used :

The sum of infinite geometric series with first term $a_1$ and common ration $r$ is calculated using the formula $S=\frac{a_1}{1-r}$ .

Here, $\left|r\right|<1$ . If $\left|r\right|\ge 1$ , then the series has no sum.

Calculation :

Consider the geometric series $-\frac{1}{8}-\frac{1}{12}-\frac{1}{18}-\frac{1}{27}+\cdots$ ,

Calculate the ratio $r$ using the formula $\frac{a_n}{a_{n-1}}$ . This ratio must hold for any value of $n$ .

  $\begin{array}{c}\frac{a_2}{a_1}=\frac{-\frac{1}{12}}{-\frac{1}{8}}\\ =\frac{8}{12}\\ =\frac{2}{3}\end{array}$

  $\begin{array}{c}\frac{a_3}{a_2}=\frac{-\frac{1}{18}}{-\frac{1}{12}}\\ =\frac{12}{8}\\ =\frac{2}{3}\end{array}$

  $\begin{array}{c}\frac{a_4}{a_3}=\frac{-\frac{1}{27}}{-\frac{1}{18}}\\ =\frac{18}{27}\\ =\frac{2}{3}\end{array}$

For the given series the value of $a_1$ is $-\frac{1}{8}$ and $r$ is $\frac{2}{3}$ .

As the common ratio $\left|\frac{2}{3}\right|<1$ , the sum of the series exists.

Then the sum of infinite geometric series is calculated using the formula $S=\frac{a_1}{1-r}$ .

  $\begin{array}{c}S=\frac{-\frac{1}{8}}{1-\frac{2}{3}}\\ =\frac{\frac{-\frac{1}{8}}{1}}{\frac{3}{8}}\\ =-\frac{3}{8}\end{array}$

Thus, the sum of geometric series $-\frac{1}{8}-\frac{1}{12}-\frac{1}{18}-\frac{1}{27}+\cdots$ is $-\frac{3}{8}$ .