Chapter 11.4, Problem 62HP

To determine

To find the value of b such that the infinite sum

  $3+9b+27b^2+81b^3+\mathrm{.....}$

Answer to Problem 62HP

  $-\frac{1}{3}<b<\frac{1}{3}$

Explanation of Solution

Given:

  $3+9b+27b^2+81b^3+\mathrm{.....}$

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:

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

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 value of b such that the infinite sum

  $3+9b+27b^2+81b^3+\mathrm{.....}$

First try to write this sum in the format ${\displaystyle {\sum }_{n=1}^{\infty }a\cdot r^{n-1}}$ .

As,

  $\begin{array}{c}3+9b+27b^2+81b^3+\mathrm{.....}=3+3\cdot 3b+3\cdot {\left(3b\right)}^2+3\cdot {\left(3b\right)}^3+\mathrm{.....}\\ ={\displaystyle {\sum }_{n=1}^{\infty }3\cdot {\left(3b\right)}^{n-1}}\end{array}$

Now, sum of infinite series ${\displaystyle {\sum }_{n=1}^{\infty }a\cdot r^{n-1}}$ exists only if $\left|r\right|<1$ .

Here,

  $\begin{array}{c}r=3b\\ \Rightarrow \left|r\right|=\left|3b\right|\end{array}$

Now,

  $\begin{array}{c}\left|3b\right|<1\\ \Rightarrow -1<3b<1\\ \Rightarrow -\frac{1}{3}<b<\frac{1}{3}\end{array}$

Thus, the sum of given series exists if and only if

  $-\frac{1}{3}<b<\frac{1}{3}$