Answer
$\dfrac{-2}{5}$
Work Step by Step
The Nth partial sum of a geometric series can be found as:
$s_n=\dfrac{a(1-r^n)}{1-r}$
The sum of a geometric series can be found as:
$S=\dfrac{a}{1-r}$
The common ratio for a geometric series is $a=\dfrac{-2}{3}, r=\dfrac{-2}{3}$;
and for a geometric series it is $|\dfrac{-2}{3}| \lt 1$; so the series converges.
Thus,
$S=\dfrac{\dfrac{-2}{3}}{1-(\dfrac{-2}{3})}=\dfrac{-2}{5}$; converges