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

$1$
The sum of a geometric series can be found as: $S=\dfrac{a}{1-r}$ The given series has first term, $a=\dfrac{3}{2}$ and common ratio $r =\dfrac{-1}{2}$ Hence, $S=\dfrac{\dfrac{3}{2}}{1-(\dfrac{-1}{2})}=1$