Chapter 12.3, Problem 62AYU

In Problems 53-68, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.

${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}8{\left(\frac{1}{3}\right)}^{k-1}}$

To determine

To verify whether the given geometric series is converges or diverges and hence find its sum accordingly.

Answer to Problem 62AYU

The given geometric series converges and its sum is 12.

Explanation of Solution

Given:

Infinite geometric series is given as ${\displaystyle \sum }_{k=1}^{\infty }8{\left(\frac{1}{3}\right)}^{k-1}=8\left(1+\frac{1}{3}+{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^3+\cdots \right)$. The first term is $a_1=1$ and the common ratio $r=\frac{a_n}{a_{n-1}}=\frac{1}{4}$.

Convergence of an infinite geometric series theorem states that, If $\left|r\right|<1$ converges. Its sum is ${\displaystyle \sum }_{k=1}^{\infty }{a}_1{r}^{k-1}=\frac{a_1}{1-r}$.

Now consider, $8\left(1+\frac{1}{3}+{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^3+\cdots \right)={\displaystyle \sum }_{k=1}^{\infty }8{\left(\frac{1}{3}\right)}^{k-1}=8.\frac{a_1}{1-r}=8\frac{1}{1-\frac{1}{3}}=12$.

Sum of the infinite geometric series is 12.