Question

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

$\sum _{k=1}^{\infty }5{\left(\frac{1}{4}\right)}^{k-1}$

Short answer

The given infinite geometric series $\sum _{k=1}^{\infty }5{\left(\frac{1}{4}\right)}^{k-1}$is convergent and its sum is $\frac{20}{3}$.

Step-by-step solution

Step 1. Write the given information.

The given geometric series is:

$\sum _{k=1}^{\infty }5{\left(\frac{1}{4}\right)}^{k-1}$

Determine the first term and common ratio.

The first term is$\left(a_1\right)=5$

The common ratio is the ratio of successive terms:

$r=\frac{1}{4}$

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 $r=\frac{1}{4}\mathrm{and}\frac{1}{4}<1$, 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$\sum _{k=1}^{\infty }{a}_1{r}^{k-1}=\frac{a_1}{1-r}$,

$$\begin{gathered} \sum _{k=1}^{\infty }5{\left(\frac{1}{4}\right)}^{k-1}=\frac{5}{1-\left({\displaystyle \frac{1}{4}}\right)} \\ =\frac{5}{{\displaystyle \frac{3}{4}}} \\ =\frac{20}{3} \end{gathered}$$

Therefore, the sum of the infinite geometric series is$\frac{20}{3}$.