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

Consider $a_n=n! e^{-n}$ Now, $l=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{(n+1)! e^{-(n+1)}}{n! e^{-n}}|$ Thus, we have $l=(\dfrac{1}{e})\lim\limits_{n \to \infty}(n+1)=(\dfrac{1}{e})(\infty)$ So, $l=\infty \gt 1$ Hence, the series Diverges absolutely by the ratio test.