Answer
Converges
Work Step by Step
Consider $a_n=(-1)^n\dfrac{n^2(n+2)}{ n! 3^{2n}}$
Now, $l=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(n+1)^2(n+1+2)}{ (n+1)! 3^{2n+2}}}{\dfrac{n^2(n+2)}{ n! 3^{2n}}}|$
Thus, we have $l=\lim\limits_{n \to \infty}|\dfrac{(n+1)^2 (n+3)}{(9) n^2 (n+1)}|=\dfrac{1}{9} \lt 1$
Hence, the series Converges absolutely by the ratio test.