Answer
$\dfrac{8}{3}$
Work Step by Step
We are given the geometric series:
$\sum_{k=1}^{\infty} 4\left(-\dfrac{1}{2}\right)^{k-1}$
Determine the elements of the geometric series:
$a_1=4\left(-\dfrac{1}{2}\right)^{1-1}=4$
$r=-\dfrac{1}{2}$
We compute $|r|$:
$|r|=\left|-\dfrac{1}{2}\right|=\dfrac{1}{2}$
Because $|r|=\dfrac{1}{2}<1$, the series converges. Determine its sum:
$S=\dfrac{a_1}{1-r}=\dfrac{4}{1-\left(-\dfrac{1}{2}\right)}=\dfrac{4}{\dfrac{3}{2}}=\dfrac{8}{3}$
