## Precalculus (6th Edition) Blitzer

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