Answer
$\dfrac{2}{3}$
Work Step by Step
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:
$S=\dfrac{a}{1-r}=\dfrac{1}{\dfrac{3}{2}}=\dfrac{2}{3}$
