Chapter 11.4, Problem 64HP

To determine

To write an infinite series with a sum that converges to 9.

Answer to Problem 64HP

  ${\displaystyle {\sum }_{n=1}^{\infty }1\cdot {\left(\frac{8}{9}\right)}^{n-1}}$

Explanation of Solution

Given:

Sum of the infinite series is 9.

Concept Used:

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:

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

Calculation:

In order to write an infinite series with a sum that converges to 9.

First let the series is

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

Then according to the formula for sum of infinite geometric series where common ratio r is $\left|r\right|<1$ :

  $\begin{array}{c}{\displaystyle {\sum }_{n=1}^{\infty }a\cdot r^{n-1}}=\frac{a}{1-r}\\ =9\left(\text{Given}\right)\end{array}$

There are many number of infinite series such that there sum converges to 9, to find one such series consider the first term $a=1$ in above formula then the common ratio can of the series witj sum converging to 9 can be find as:

  $\begin{array}{c}{\displaystyle {\sum }_{n=1}^{\infty }1\cdot r^{n-1}}=9\\ \frac{1}{1-r}=9\\ \frac{1}{9}=1-r\\ r=1-\frac{1}{9}\\ r=\frac{9-1}{9}\\ r=\frac{8}{9}\end{array}$

Thus, one example of an infinite series with sum converging to 9

  ${\displaystyle {\sum }_{n=1}^{\infty }1\cdot {\left(\frac{8}{9}\right)}^{n-1}}$ .