Chapter 12, Problem 41RE

To determine

To find: Whether the geometric series converges or diverges. Find the sum if the series converges.

Answer to Problem 41RE

The infinite series is convergent and its sum is $\frac{9}{2}$ .

Explanation of Solution

Given:

The given series is $3+1+\frac{1}{3}+\frac{1}{9}$

Calculation:

Consider given series is $3+1+\frac{1}{3}+\frac{1}{9}$ .

The common ratio of the series is,

  $r=\frac{1}{3}$

Consider the formula for the geometric series is,

  ${\displaystyle \sum _{k=1}^{\infty }{a}_1{r}^{k-1}}=\frac{a_1}{1-r}$

The above series is convergent when $r<1$ as $\frac{1}{3}<1$ is true the series is convergent.

Then, the sum is,

  $\begin{array}{c}{\displaystyle \sum _{k=1}^{\infty }\left(3\right){\left(\frac{1}{3}\right)}^{k-1}}=\frac{3}{1-\frac{1}{3}}\\ =\frac{9}{2}\end{array}$

Thus, the infinite series is convergent and its sum is $\frac{9}{2}$ .