Answer
Converges Absolutely
Work Step by Step
Let $a_n=\dfrac{(-3)^n}{n!}$ and $a_{n+1}=\dfrac{(-3)^{n+1}}{(n+1)!}$
Apply Ratio Test.
$\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty} |\dfrac{\dfrac{(-3)^{n+1}}{n!(n+1)}}{\dfrac{(-3)^n}{n!}}|$
$\implies \lim\limits_{n \to \infty} \dfrac{3}{(n+1)}=\lim\limits_{n \to \infty} \dfrac{\dfrac{3}{n}}{(1+\dfrac{1}{n})}$
Thus, we have $\dfrac{3}{\infty}=0 \lt 1$
Hence, the series Converges Absolutely by Ratio Test.
