## Calculus 8th Edition

According to Ratio Test The sequence $a_{n}=\frac{1}{n!}$ will be converging if $\lim\limits_{n \to \infty} |\frac{a_{n+1}}{a_{n}}| \lt 1$ $\frac{a_{n+1}}{a_{n}}=\frac{\frac{1}{(_n+1)!}}{\frac{1}{n!}}$ $=\frac{n!}{(n+1)!}$ $=\frac{n!}{n! \times (n+1)}$ $=\frac{1}{n+1}$ $\lim\limits_{n \to \infty} |\frac{a_{n+1}}{a_{n}}| =\lim\limits_{n \to \infty} \frac{1}{n+1}=0$ Therefore, the given sequence is converging. Alternate Solution: Clearly $n!=n (n-1)(n-2)...(3)(2)\geq 2.2.2....2.2=2^{n-1}$ Thus, $\frac{1}{n!} \leq \frac{1}{2^{n-1}} \Sigma_{n=1}^{\infty } \frac{1}{2^{n-1}}$ is a convergent geometric series $|r| =\frac{1}{2}\lt 1$, so , $\Sigma_{n=1}^{\infty } \frac{1}{n!}$ converges by the Comparison Test.