Chapter 12, Problem 47RE

To determine

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

Answer to Problem 47RE

The infinite series converges and its sum is $8$ .

Explanation of Solution

Given:

The given series is ${\displaystyle \sum _{k=1}^{\infty }4{\left(\frac{1}{2}\right)}^{k-1}}$ .

Calculation:

Consider the formula for the geometric series is,

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

Consider given series is,

  ${\displaystyle \sum _{k=1}^{\infty }4{\left(\frac{1}{2}\right)}^{k-1}}$

The above series is convergent when $\left|r\right|<1$ as $\left|\frac{1}{2}\right|<1$ is true the series converges.

Then, the sum is,

  $\begin{array}{c}{\displaystyle \sum _{k=1}^{\infty }4{\left(\frac{1}{2}\right)}^{k-1}}=\frac{4}{1-\frac{1}{2}}\\ =8\end{array}$

Thus, the infinite series converges and its sum is $8$ .