Question

In Problems 22–25, determine whether each infinite geometric series converges or diverges. If it converges, find its sum.

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

Short answer

The geometric series converges.

Its sum is:$8$

Step-by-step solution

Step 1. Given information.

Geometric series:

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

Step 2. To find common ratio and first term.   

By comparing $\sum _{k=1}^{\infty }4{\left(\frac{1}{2}\right)}^{k-1}$to $\sum _{k=1}^{\infty }a.r^{k-1}$

We get first term:$a=4$

Common ratio:$r=\frac{1}{2}$

Since common ratio is less than 1.

Hence the series converges.

Step 3. To find sum of infinite terms.

$\begin{array}{rcl}{S}_{\infty }& =& \frac{4}{1-{\displaystyle \frac{1}{2}}}\\ & =& \frac{4}{{\displaystyle \frac{1}{2}}}\\ & =& 4\times 2\\ & =& 8\\ & & \end{array}$