Chapter 12.3, Problem 51AYU

To determine

To find: Whether the geometric series converges or diverges and find the sum if converges.

Answer to Problem 51AYU

The series is convergent and the sum of series is $\frac{3}{2}$ .

Explanation of Solution

Given information:

The series is $1+\frac{1}{3}+\frac{1}{9}+\mathrm{........}$ .

Calculation:

For the given series $a_1=1$ , $a_2=\frac{1}{3}$ and $a_3=\frac{1}{9}$ .

Calculate $\frac{a_2}{a_1}$ .

  $\begin{array}{c}\frac{a_2}{a_1}=\frac{\frac{1}{3}}{1}\\ =\frac{1}{3}\end{array}$

Calculate $\frac{a_3}{a_1}$ .

  $\begin{array}{c}\frac{a_3}{a_1}=\frac{\frac{1}{9}}{\frac{1}{3}}\\ =\frac{1}{3}\end{array}$

Since the ratio of successive terms is same it is a geometric series with common ratio $r=\frac{1}{3}$ .

Since $\left|r\right|<1$ the series is convergent.

The sum of infinite series which is convergent is given by,

  $\begin{array}{c}{S}_{\infty }=\frac{a}{1-r}\\ =\frac{1}{1-\frac{1}{3}}\\ =\frac{1}{\frac{2}{3}}\\ =\frac{3}{2}\end{array}$

Therefore, the series is convergent and the sum of series is $\frac{3}{2}$ .