Answer
Convergent
Sum: $8$
Work Step by Step
We are given the geometric series:
$\sum_{k=1}^{\infty} 4\left(\dfrac{1}{2}\right)^{k-1}$
The elements of the geometric series are:
$a_1=4\left(\dfrac{1}{2}\right)^{1-1}=4$
$r=\dfrac{1}{2}$
Compute $|r|$:
$|r|=\left|\dfrac{1}{2}\right|=\dfrac{1}{2}$
Because $|r|=\dfrac{1}{2}<1$, the series converges.
Determine its sum:
$S=\dfrac{a_1}{1-r}$
$S=\sum_{k=1}^{\infty}4\left(\dfrac{1}{2}\right)^{k-1}=\dfrac{4}{1-\dfrac{1}{2}}=\dfrac{4}{\dfrac{1}{2}}=8$
