Chapter 12.3, Problem 56AYU

To determine

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

Answer to Problem 56AYU

The series is convergent and the sum of series is $\frac{4}{7}$ .

Explanation of Solution

Given information:

The series is $1-\frac{3}{4}+\frac{9}{16}-\frac{27}{64}+\mathrm{......}$ .

Calculation:

For the given series $a_1=1$ , $a_2=-\frac{3}{4}$ and $a_3=\frac{9}{16}$ .

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

  $\begin{array}{c}\frac{a_2}{a_1}=\frac{-\frac{3}{4}}{1}\\ =-\frac{3}{4}\end{array}$

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

  $\begin{array}{c}\frac{a_3}{a_1}=\frac{\frac{9}{16}}{\frac{-3}{4}}\\ =-\frac{3}{4}\end{array}$

Since the ratio of successive terms is same it is a geometric series with common ratio $r=-\frac{3}{4}$ .

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

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

  $\begin{array}{c}{S}_{\infty }=\frac{a_1}{1-r}\\ =\frac{1}{1-\left(-\frac{3}{4}\right)}\\ =\frac{1}{\frac{7}{4}}\\ =\frac{4}{7}\end{array}$

Therefore, the series is convergent and the sum of series is $\frac{4}{7}$ .