Answer
$\dfrac{18}{5}$
Work Step by Step
We are given the geometric series:
$\sum_{k=1}^{\infty} 6\left(-\dfrac{2}{3}\right)^{k-1}$
Determine the elements of the geometric series:
$a_1=6\left(-\dfrac{2}{3}\right)^{1-1}=6$
$r=-\dfrac{2}{3}$
We compute $|r|$:
$|r|=\left|-\dfrac{2}{3}\right|=\dfrac{2}{3}$
Because $|r|=\dfrac{2}{3}<1$, the series converges. Determine its sum:
$S=\dfrac{a_1}{1-r}=\dfrac{6}{1-\left(-\dfrac{2}{3}\right)}=\dfrac{6}{\dfrac{5}{3}}=\dfrac{18}{5}$
