Answer
$a_{n}$ converges to $0$.
Work Step by Step
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 the 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 $0$.
