## Calculus: Early Transcendentals 8th Edition

Let $a_n=\frac{n!}{100^n}$. Then $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|= \lim\limits_{n \to \infty}\left|\frac{\frac{(n+1)!}{100^{n+1}}}{\frac{n!}{100^n}}\right|=\lim\limits_{n \to \infty}\left|\frac{(n+1)!}{100^{n+1}}\cdot\frac{100^n}{n!}\right|\lim\limits_{n \to \infty}\left|\frac{(n+1)!}{n!}\cdot\frac{100^n}{100^{n+1}}\right|=\lim\limits_{n \to \infty}\left|(n+1)\cdot\frac{1}{100}\right| =$ $$\frac{1}{100}\cdot\lim\limits_{n \to \infty}\left|n+1\right| =\infty.$$ Since $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|=\infty$, the series $\displaystyle{\sum_{n=1}^{\infty}\frac{n!}{100^n}}$ is divergent by the Ratio Test.