Answer
Converges
Work Step by Step
Consider $a_n=\dfrac{(n+3)!}{3! n! (3^n)}$
Now, $l=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{((n+1)+3)!}{3! (n+1)! (3^{n+1})}}{\dfrac{(n+3)!}{3! n! (3^n)}}|$
Thus, we have $l=(\dfrac{1}{3})\lim\limits_{n \to \infty}(\dfrac{n+4}{n+1})=(\dfrac{1}{3})(1)$
So, $l=\dfrac{1}{3} \lt 1$
Hence, the series Converges by the ratio test.