Chapter 12.3, Problem 63AYU

To determine

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

Answer to Problem 63AYU

The series is convergent and the sum of series is $\frac{18}{5}$ .

Explanation of Solution

Given information:

The series is ${\displaystyle \sum _{k=1}^{\infty }6{\left(-\frac{2}{3}\right)}^{k-1}}$ .

Calculation:

For the given series $a_1=6$ , and $r=-\frac{2}{3}$

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

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

  $\begin{array}{c}{S}_{\infty }=\frac{a_1}{1-\left(r\right)}\\ =\frac{6}{1-\left(-\frac{2}{3}\right)}\\ =\frac{6}{\frac{5}{3}}\\ =\frac{18}{5}\end{array}$

Therefore, the series is convergent and the sum of series is $\frac{18}{5}$ .