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

Consider $a_n=\dfrac{(-3)^n}{n!}$ By the Ratio Test: $l=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty} |\dfrac{\dfrac{(-3)^{n+1}}{(n+1)!}}{\dfrac{(-3)^n}{n!}}|$ or, $=\lim\limits_{n \to \infty} |\dfrac{\dfrac{(-3)^{n+1}}{n!(n+1)}}{\dfrac{(-3)^n}{n!}}|$ or, $\lim\limits_{n \to \infty} \dfrac{3}{(n+1)}=\lim\limits_{n \to \infty} \dfrac{3/n}{(1+1/n)}$ or, $\dfrac{3}{\infty}=0$ so, $l=0 \lt 1$ Thus, the series Converges Absolutely by the Ratio Test.