Answer
$\dfrac{3}{8}$
Work Step by Step
Here, we have $\sum_{i=0}^4\dfrac{(-1)^i}{i!}=\dfrac{(-1)^0}{0!}+\dfrac{(-1)^1}{1!}+\dfrac{(-1)^2}{2!}+\dfrac{(-1)^3}{3!}+\dfrac{(-1)^4}{4!}$
$=\dfrac{1}{1}+(\dfrac{-1}{1})+\dfrac{1}{2}+(\dfrac{-1}{6})+(\dfrac{1}{24})$
$=\dfrac{12}{24}-\dfrac{4}{24}+\dfrac{1}{24}$
$\dfrac{9}{24}$
$=\dfrac{3}{8}$
