Question

In Problems 51–66, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.

$6+2+\frac{2}{3}+.....$

Short answer

The given infinite geometric series $6+2+\frac{2}{3}+.....$is convergent and its sum is 9.

Step-by-step solution

Step 1. Write the given information.

The given geometric series is:

$6+2+\frac{2}{3}+.....$

Step 2. Find the common ratio.

$a_1=6,a_2=2,a_3=\frac{2}{3}$

The common ratio is the ratio of successive terms:

$$\begin{gathered} \frac{2}{6}=\frac{1}{3},\frac{\frac{2}{3}}{2}=\frac{1}{3} \\ r=\frac{1}{3} \end{gathered}$$

Step 3. Determine whether the infinite geometric series is converges or diverges.

If$\left|r\right|<1$then the infinite geometric series$\sum _{k=1}^{\infty }{a}_1{r}^{k-1}$converges.

As we can see that$r=\frac{1}{3}\mathrm{and}\frac{1}{3}<1$, therefore the given series is convergent.

Step 4. Find the sum of the series.

Use the formula to find the sum of the given geometric series $\sum _{k=1}^{\infty }{a}_1{r}^{k-1}=\frac{a_1}{1-r}$,

$$\begin{gathered} \sum _{k=1}^{\infty }6·{\left(\frac{1}{3}\right)}^{k-1}=\frac{6}{\left(1-{\displaystyle \frac{1}{3}}\right)} \\ =\frac{6}{{\displaystyle \frac{2}{3}}} \\ =9 \end{gathered}$$

Therefore, the sum of the infinite geometric series is 9.