Chapter 10.1, Problem 11E

To determine

Check whether the series converges or diverges. If converges, then find sum.

Answer to Problem 11E

The series converges and the sum is $3$ .

Explanation of Solution

Given Information:

The series is $1+\frac{2}{3}+{\left(\frac{2}{3}\right)}^2+{\left(\frac{2}{3}\right)}^3+\cdots +{\left(\frac{2}{3}\right)}^n+\cdots$ .

Formula Used:

The sum of the geometric series when ration is less than 1:

  $S=\frac{a}{1-r}$ , where $a$ is initial term and $r$ is the common ratio.

Calculation:

The series is $1+\frac{2}{3}+{\left(\frac{2}{3}\right)}^2+{\left(\frac{2}{3}\right)}^3+\cdots +{\left(\frac{2}{3}\right)}^n+\cdots$ .

Rewrite the series as:

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

In the series ${\displaystyle \sum _{n=1}^{\infty }{\left(\frac{2}{3}\right)}^n}$ , $a=1$ and $r=\frac{2}{3}$ .

The ratio of the geometric series is less than 1 so, the series converges.

Find the sum of the series.

Substitute $1$ for $a$ and $\frac{2}{3}$ for $r$ in the formula $\frac{a}{1-r}$ .

  $\begin{array}{c}S=\frac{1}{1-\frac{2}{3}}\\ =\frac{1}{\frac{3-2}{3}}\\ =3\end{array}$

Hence, the series converges and the sum is $3$ .