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

$\dfrac{2}{3}$
Here, $r=\dfrac{-1}{2}, a=1$ The Nth partial sum of a geometric series can be found as: $s_n=\dfrac{a(1-r^n)}{1-r}=\dfrac{1(1-(\dfrac{-1}{2})^n)}{1-(\dfrac{-1}{2}})$ The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}=\dfrac{1}{\dfrac{3}{2}}=\dfrac{2}{3}$