Chapter 11.4, Problem 29PPS

To determine

To find the sum of infinite series(if it exists)

  ${\displaystyle {\sum }_{k=1}^{\infty }\frac{4}{3}\cdot {\left(\frac{5}{4}\right)}^{k-1}}$

Answer to Problem 29PPS

The sum doesn’t exist.

Explanation of Solution

Given:

  ${\displaystyle {\sum }_{k=1}^{\infty }5\cdot 4^{k-1}}$

Concept Used:

1. Sum of an infinite geometric series $a+a\cdot r+a\cdot r^2+a\cdot r^3\mathrm{......}$ , whose first term is a and common ratio is r such that $\left|r\right|<1$is:

  ${\displaystyle {\sum }_{n=1}^{\infty }a\cdot r^{n-1}}=\frac{a}{1-r}$

2. Sum of an infinite geometric series $a+a\cdot r+a\cdot r^2+a\cdot r^3\mathrm{......}$ , whose first term is a and common ratio is r such that $r>1$ is divergent(doesn’t exist).

Calculation:

In order to find the sum of infinite series

  ${\displaystyle {\sum }_{k=1}^{\infty }\frac{4}{3}\cdot {\left(\frac{5}{4}\right)}^{k-1}}$

First check whether this sum exists or not, for that check if the common ratio of the series satisfies $\left|r\right|<1$ or $r>1$ .

Comparing the given series with

  $\begin{array}{c}{\displaystyle {\sum }_{n=1}^{\infty }a\cdot r^{n-1}}\\ \Rightarrow r=\frac{5}{4}\\ \Rightarrow r=1.25\end{array}$ ,

Since, the common ratio of this infinite series is greater than 1.

Thus, the given sum of is divergent.

Thus, the sum doesn’t exist.