Chapter 9.4, Problem 43E

Solution

To identify: If the series $3^{-1}+3^{-2}+3^{-3}+\cdots +3^{-n}+\cdots$ converges or diverges.

The options are:

  $\left(a\right)$ Converges to $\frac{1}{2}$

  $\left(b\right)$ converges to $\frac{1}{3}$

  $\left(c\right)$ converges to $\frac{2}{3}$

  $\left(d\right)$ converges to $\frac{3}{2}$

  $\left(e\right)$ diverges

The correct option is $\left(a\right)$ Converges to $\frac{1}{2}$ .

Given information:

The given series is $3^{-1}+3^{-2}+3^{-3}+\cdots +3^{-n}+\cdots$ .

Formula used:

A geometric series ${\displaystyle {\sum }_{k=1}^{\infty }a\cdot r^{k-1}}$ , converges if and only if $\left|r\right|<1$ .If it does converge, the sum: $S=\frac{a_1}{1-r}$ .

Calculation:

Consider the geometric series,

  $3^{-1}+3^{-2}+3^{-3}+\cdots +3^{-n}+\cdots$

It can also be written as: $\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots +\frac{1}{3^n}+\cdots$

The common ratio ( r ) is:

  $r=\frac{\frac{1}{9}}{\frac{1}{3}},r=\frac{1}{3}$

For the given series the value of $a_1$ is $\frac{1}{3}$ and $r\text{is}\frac{1}{3}$ . As the common ratio $\left|r\right|<1$ , then the given series converges.

Now find the sum using the sum formula $S=\frac{a_1}{1-r}$ :

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

Therefore, the given series converges to $\frac{1}{2}$ . Option $\left(a\right)$ Converges to $\frac{1}{2}$ should be the correct answer.