## Calculus: Early Transcendentals 8th Edition

(a) By the ratio test, the series is convergent. (b) $\lim\limits_{n \to \infty}\frac{x^{n}}{n!}=0$
(a) $\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{x^{n+1}}{(n+1)!}}{\dfrac{x^{n}}{n!}}|$ $\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{x^{n}}{(n)!}\times \frac{x}{(n+1)}}{\dfrac{x^{n}}{n!}}|$ $=\lim\limits_{n \to \infty}|\frac{{a_{n}\times \frac{x}{(n+1)}}}{a_{n}}|$ $=\lim\limits_{n \to \infty}|\frac{x}{(n+1)}|$ $=0\lt 1$ By the ratio test, the series is convergent. (b) From part (a), we know that $\Sigma a_{n}=\sum_{n=1}^{\infty}\frac{x^{n}}{n!}$ is a convergent series. Thus, $\lim\limits_{n \to \infty} a_{n}=\lim\limits_{n \to \infty}\frac{x^{n}}{n!}=0$