What is the convergence of this series? I am trying to find the convergence of $\sum_{n=1}^\infty \frac{(n!)^3}{(3n)!}$. I proceeded with the ratio test to get $$\lim_{n\to \infty}\frac{(n+1!)^3}{(3(n+1))!} \cdot \frac{(3n)!}{(n!)^3} = \lim_{n\to \infty}\frac{(n+1)^3}{3(n+1)} = \lim_{n\to \infty}\frac{(n+1)^2}{3}=\infty $$. So, the ratio test says that this series diverges, but from what I have looked up, this series should converge. Is there something wrong in my calculations? Any help would be greatly appreciated. Thanks!

As pointed out in the comments, there is an algebra error that is easily fixed and convergence follows. For an alternative proof, write out
\begin{align*}
(3n)! &= \big(1 \cdot 2 \cdots n\big)\big((n + 1) \cdot (n + 2) \cdots 2n)\big) \big((2n + 1) \cdot (2n + 2) \cdots 3n\big) \\
&\ge n! \cdot (2^n n!) \cdot (3^n n!) \\
&= 6^n (n!)^3
\end{align*}
The inequality follows from the fact that $n + k \ge 2k$ for each $k \in [1, n]$ (giving $2^n$) and something similar for the last term. Hence, the series is bounded by the convergent geometric series $\sum_{n =1 }^{\infty} (1/6)^n$, which even gives rather rapid convergence.
