Answer
$6$
Work Step by Step
We are given the geometric series:
$\sum_{k=1}^{\infty} 3\left(\dfrac{2}{3}\right)^{k}$
Determine the elements of the geometric series:
$a_1=3\left(\dfrac{2}{3}\right)^{1}=2$
$r=\dfrac{2}{3}$
We compute $|r|$:
$|r|=\left|\dfrac{2}{3}\right|=\dfrac{2}{3}$
Because $|r|=\dfrac{2}{3}<1$, the series converges. Determine its sum:
$S=\dfrac{a_1}{1-r}=\dfrac{2}{1-\dfrac{2}{3}}=\dfrac{2}{\dfrac{1}{3}}=6$
