Answer
convergent
Work Step by Step
Given
$$\sum_{n=0}^{\infty} \frac{(n !)^{2}}{(3 n) !}$$
Here
\begin{aligned}
a_n&= \frac{(n !)^{2}}{(3 n) !}\\
a_{n+1}&= \frac{((n+1) !)^{2}}{(3 n+3) !}
\end{aligned}
Apply Ratio Test
\begin{aligned}
\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|&=
\lim _{n \rightarrow \infty}\left| \frac{((n+1) !)^{2}}{(3 n+3) !} \frac{(3 n) !} {(n !)^{2}}\right|\\
&= \lim _{n \rightarrow \infty}\left| \frac{(n+1)^2(n !)^{2}}{(3 n+3)(3n+2)(3n+1)(3n) !} \frac{(3 n) !} {(n !)^{2}}\right|\\
&= \lim _{n \rightarrow \infty}\left| \frac{(n+1)^2 }{(3 n+3)(3n+2)(3n+1) } \right|\\
&= \lim _{n \rightarrow \infty}\left|\frac{n+1}{3\left(3n+2\right)\left(3n+1\right)} \right|\\
&= 0<1
\end{aligned}
Then the given series is convergent
