## Precalculus (10th Edition)

$6$
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: $S=\dfrac{a_1}{1-r}=\dfrac{2}{1-\dfrac{2}{3}}=\dfrac{2}{\dfrac{1}{3}}=6$