Answer
$\displaystyle \sum_{j=0}^{5}\frac{(-1)^j}{j!}=\frac{17}{60}$
Work Step by Step
$\displaystyle \sum_{j=0}^{5}\frac{(-1)^j}{j!}=\frac{(-1)^0}{0!}+\frac{(-1)^1}{1!}+\frac{(-1)^2}{2!}+\frac{(-1)^3}{3!}+\frac{(-1)^4}{4!}+\frac{(-1)^5}{5!}=1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}=\frac{60-20+5-1}{120}=\frac{34}{120}=\frac{17}{60}$
