Chapter 9, Problem 9CT

To determine

To calculate: The value of the expression ${{\displaystyle \sum _{n=1}^{\infty }\left(-\frac{1}{3}\right)}}^n$ .

Answer to Problem 9CT

The value of the expression ${{\displaystyle \sum _{n=1}^{\infty }\left(-\frac{1}{3}\right)}}^n$ is $\frac{-1}{4}$ .

Explanation of Solution

Given information:

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

Formula used:

The sum of infiniteterms of the geometric sequence is denoted by $S_n=\frac{a}{1-r}$ , where a is the first term andr is the common ratio.

Calculation:

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

Expand the expression as,

  ${{\displaystyle \sum _{n=1}^{\infty }\left(-\frac{1}{3}\right)}}^n={\left(-\frac{1}{3}\right)}^1+{\left(-\frac{1}{3}\right)}^2+{\left(-\frac{1}{3}\right)}^3+\dots +{\left(-\frac{1}{3}\right)}^n+\dots$

Here first term is $a=\left(-\frac{1}{3}\right)$ and common ration is

  $\begin{array}{c}r=\frac{a_2}{a_1}\\ =\frac{{\left(-\frac{1}{3}\right)}^2}{\left(-\frac{1}{3}\right)}\\ =\left(-\frac{1}{3}\right)\end{array}$

Similarly,

  $\begin{array}{c}r=\frac{a_2}{a_1}\\ =\frac{a_3}{a_2}\\ ⋮\\ =\left(-\frac{1}{3}\right)\end{array}$

The above sequence forms a geometric sequence since common ration between each consecutive term is same.

The sum of infinite terms of the geometric sequence is denoted by $S_n=\frac{a}{1-r}$ , where a is the first term and r is the common ratio.

Therefore, substitute a is $\left(-\frac{1}{3}\right)$ and r as $\left(-\frac{1}{3}\right)$ in the above expression,

  $\begin{array}{c}{S}_n=\frac{\left(-\frac{1}{3}\right)}{1-\left(-\frac{1}{3}\right)}\\ =\frac{\frac{-1}{3}}{1+\frac{1}{3}}\\ =\frac{\frac{-1}{3}}{\frac{4}{3}}\\ =\frac{-1}{4}\end{array}$

Thus, the value of the expression ${{\displaystyle \sum _{n=1}^{\infty }\left(-\frac{1}{3}\right)}}^n$ is $\frac{-1}{4}$ .