## Thomas' Calculus 13th Edition

As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists. Consider $a_n= \dfrac{(n+1)!}{n!}$ or, $a_n= \dfrac{n!(n+1)}{n!}=(n+1)$ Apply limits to both sides. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty} (n+1)$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}n+\lim\limits_{n \to \infty}(1)$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty} (n+1)=\infty$ Hence, the sequence is divergent.