## Calculus: Early Transcendentals 8th Edition

Ratio Test: $\lim\limits_{n \to +\infty}|\frac{a_{n}+1}{a_{n}}|=\lim\limits_{n \to +\infty}|\frac{1}{n+1!}\times \frac{n!}{1}|$ Here, $\frac{1}{n+1!}$ can also written as $\frac{1}{n+1!}=\frac{1}{n!}\times \frac{1}{n+1}$ Therefore, $\lim\limits_{n \to +\infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to +\infty}|\frac{1}{n+1}|=0$ So, the given series converges by Ratio Test when limit equals $0$ which is $\lt 1$ . Hence, the statement is TRUE.