## Calculus 8th Edition

$a_{n}$ converges to $0$.
Use Ratio Test $\Sigma_{n=1}^{\infty}$ $a_{n}=\Sigma_{n=1}^{\infty} \frac{(-3)^{n}}{n!}$ $|a_{n}|=\frac{3^{n}}{n!}$ $\lim\limits_{n \to \infty} |\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty} |\frac{3^{n+1}}{(n+1)!}\times \frac{n!}{3^{n}}|= \lim\limits_{n \to \infty} |\frac{3n!}{(n+1)!}|$ By definition $(n+1)!=(n+1)n!$ $\lim\limits_{n \to \infty} |\frac{3n!}{(n+1)n!}| = \lim\limits_{n \to \infty} |\frac{3}{n+1}| = \frac{3}{\infty}= 0 \lt 1$ Thus, the series $\Sigma|a_{n}|$ is convergent by ratio test and by theorem $\lim\limits_{n \to \infty} |a_{n}|=0$ leads to $\lim\limits_{n \to \infty} a_{n}=0$ Therefore, $a_{n}$ converges to $0$.