Answer
$6$
Work Step by Step
We are given the geometric series:
$\sum_{k=1}^{\infty} 2\left(\dfrac{3}{4}\right)^{k}$
Determine the elements of the geometric series:
$a_1=2\left(\dfrac{3}{4}\right)^{1}=\dfrac{3}{2}$
$r=\dfrac{3}{4}$
We compute $|r|$:
$|r|=\left|\dfrac{3}{4}\right|=\dfrac{3}{4}$
Because $|r|=\dfrac{3}{4}<1$, the series converges. Determine its sum:
$S=\dfrac{a_1}{1-r}=\dfrac{\dfrac{3}{2}}{1-\dfrac{3}{4}}=\dfrac{\dfrac{3}{2}}{\dfrac{1}{4}}=6$
