Answer
$12$
Work Step by Step
We are given the geometric series:
$\sum_{k=1}^{\infty} 8\left(\dfrac{1}{3}\right)^{k-1}$
Determine the elements of the geometric series:
$a_1=8\left(\dfrac{1}{3}\right)^{1-1}=8$
$r=\dfrac{1}{3}$
We compute $|r|$:
$|r|=\left|\dfrac{1}{3}\right|=\dfrac{1}{3}$
Because $|r|=\dfrac{1}{3}<1$, the series converges. Determine its sum:
$S=\dfrac{a_1}{1-r}==\dfrac{8}{1-\dfrac{1}{3}}=\dfrac{8}{\dfrac{2}{3}}=12$