Chapter 11.4, Problem 24PPS

To determine

To find the sum of the infinite series (if it exists):

  $\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\mathrm{......}$

Answer to Problem 24PPS

  $0.52$

Explanation of Solution

Given:

  $\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\mathrm{......}$

Concept Used:

1. Sum of an infinite series $a+a\cdot r+a\cdot r^2+a\cdot r^3\mathrm{......}$ , whose first term is a and common ratio is r such that $\left|r\right|<1$is:

  ${\displaystyle {\sum }_{n=0}^{\infty }a\cdot r^n}=\frac{a}{1-r}$

2. Sum of an infinite series $a+a\cdot r+a\cdot r^2+a\cdot r^3\mathrm{......}$ , whose first term is a and common ratio is r such that $r>1$ does not exist.

Calculation:

In order to find the sum of infinite series

  $\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\mathrm{......}$

Use the formula

  ${\displaystyle {\sum }_{n=0}^{\infty }a\cdot r^n}=\frac{a}{1-r}$

Note from the given infinite series that first term is $\frac{1}{2}$ , and the common ratio ( r ) can be find by dividing second term by first term, as

  $r=\frac{a_2}{a_1}$

Since, here

  $a_2=\frac{1}{6},a_1=\frac{1}{2}$

Thus,

  $\begin{array}{c}r=\frac{\frac{1}{6}}{\frac{1}{2}}\\ =\frac{1}{6}\times \frac{2}{1}\\ =\frac{1}{3}\\ =0.33\end{array}$

Since, $\left|r\right|<1$ .

Thus, sum of given infinite series can be find using the formula

  ${\displaystyle {\sum }_{n=0}^{\infty }a\cdot r^n}=\frac{a}{1-r}$

As

  $\begin{array}{c}\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\mathrm{......}=\frac{\frac{1}{2}}{1-0.33}\\ =\frac{0.5}{0.96}\\ =0.52\end{array}$

Thus, the sum of the infinite series,

  $\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\mathrm{......}=0.52$