## University Calculus: Early Transcendentals (3rd Edition)

$e^{-x}=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^n}{n!}$
Since, we know that the Maclaurin Series for $e^x$ is defined as: $e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}$ We need to replace in the above series $x$ with $-x$. we have $e^{-x}=\Sigma_{n=0}^\infty \dfrac{(-x)^n}{n!}$ or, $e^{-x}=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^n}{n!}$