Chapter 8.3, Problem 74E

To determine

To find the sum of the infinite geometric series ${\displaystyle {\sum }_{n=1}^{\infty }8{\left(\frac{5}{3}\right)}^{n-1}}$ .

Answer to Problem 74E

  ${\displaystyle {\sum }_{n=1}^{\infty }8{\left(\frac{5}{3}\right)}^{n-1}}$ diverges.

Explanation of Solution

Given information :

The infinite geometric series is defined as ${\displaystyle {\sum }_{n=1}^{\infty }8{\left(\frac{5}{3}\right)}^{n-1}}$ .

Concept used:

In a sequence $a_1,a_2,a_3,a_4,\dots \dots ,a_n,\dots$ if ratio of the consecutive terms is the same then it is defined as Geometric Sequence.

So, a number r is defined as the common ratio of the geometric sequence if $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\dots \dots =r.$

The sum of an infinite geometric series $a_1+a_{1}r+a_1{r}^2+a_1{r}^3+\cdots +a_1{r}^{n-1}+\cdots$ is $S={\displaystyle {\sum }_{i=0}^{\infty }{a}_1{r}^i=\frac{a_1}{1-r}}$ where $\left|r\right|<1$ .

Calculations:

Consider the given infinite geometric series.

  $\begin{array}{c}{\displaystyle {\sum }_{n=1}^{\infty }8{\left(\frac{5}{3}\right)}^{n-1}}=8\left[{\left(\frac{5}{3}\right)}^0+{\left(\frac{5}{3}\right)}^1+{\left(\frac{5}{3}\right)}^2+\cdots +{\left(\frac{5}{3}\right)}^n+\cdots \right]\\ =8\left[1+\frac{5}{3}+{\left(\frac{5}{3}\right)}^2+\cdots +{\left(\frac{5}{3}\right)}^n+\cdots \right]\end{array}$

Here, $a_1=8$ and common ratio $r=\frac{a_2}{a_1}=\frac{5\left(\frac{8}{3}\right)}{5}=\frac{8}{3}$ .

Now, here $\left|r\right|=\frac{5}{3}>1$

So, the sum is divergent.

It is not possible to find out the sum of the infinite series.