## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{-2}{5}$
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