Answer
$\dfrac{1}{2}$
Work Step by Step
Here, we have $a_n= a_1 r^{n-1}$ for the Geometric series.
First term $a_1= \dfrac{2}{3}$ and Common ratio $r=\dfrac{-1}{3}$
The sum of an infinite Geometric Series can be found using: $S_n=\dfrac{a_1}{1-r}$
Thus, $S_n=\dfrac{ \dfrac{2}{3}}{1-( \dfrac{-1}{3})}=\dfrac{ \dfrac{2}{3}}{1+ \dfrac{1}{3}}$
Hence, $S_n=\dfrac{1}{2}$