Chapter 11.4, Problem 7CYU

To determine

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

  $\frac{1}{4}+\frac{3}{8}+\frac{9}{16}+\mathrm{.....}$

Answer to Problem 7CYU

The sum doesn’t exist.

Explanation of Solution

Given:

  $\frac{1}{4}+\frac{3}{8}+\frac{9}{16}+\mathrm{.....}$

Concept Used:

 1. Sum of an infinite geometric 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 geometric 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 $r>1$ is divergent(doesn’t exist).

Calculation:

In order to find the sum of infinite series

  $\frac{1}{4}+\frac{3}{8}+\frac{9}{16}+\mathrm{.....}$

First find its first term and common ratio.

Note from the given geometric series note that first term is $\frac{1}{4}$ , and the common ratio ( r ) can be find by dividing second term by first term, as

  $r=\frac{a_2}{a_1}$

Since, here

  $a_2=\frac{3}{8},a_1=\frac{1}{4}$

Thus,

  $\begin{array}{c}r=\frac{\frac{3}{8}}{\frac{1}{4}}\\ =\frac{3}{8}\times \frac{4}{1}\\ =\frac{3}{2}\\ =1.5\\ >1\end{array}$

Since, the common ratio of the series is greater than 1.

Thus, the given sum of is divergent.

Thus, the sum doesn’t exist.