## Calculus 8th Edition

Since, $e^{x}=\Sigma_{n=0}^{\infty}\frac{x^{n}}{n!}$ Put the value of $x$ into the series as $\frac{1}{e}=e^{-1}$, so we can plug in $x=-1$ Therefore, $e^{-1}=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}}{n!}$ OR $\frac{1}{e}=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}}{n!}$ Hence, the statement is TRUE.