Answer
$\dfrac{4}{5}$
Work Step by Step
Here, $r=\dfrac{-1}{4}, a=1$
The Nth partial sum of a geometric series can be found as:
$s_n=\dfrac{a(1-r^n)}{1-r}=\dfrac{1(1-(\dfrac{-1}{4})^n)}{1-(\dfrac{-1}{4}})$
The sum of a geometric series can be found as:
$S=\dfrac{a}{1-r}=\dfrac{1}{\dfrac{5}{4}}=\dfrac{4}{5}$