Find the interval of convergence $\sum\limits_{n=1}^\infty \frac{(3n)!(x^n)}{(n!)^3}$ $\displaystyle \sum_{n=1}^\infty \dfrac{(3n)!(x^n)}{(n!)^3}$
I already found the interval of convergence to be $\displaystyle -\frac{1}{27} < x < \frac{1}{27}$. I am having trouble checking the endpoint of $\displaystyle x= \frac{1}{27}$. I need to figure out of $\displaystyle \sum_{n=1}^\infty \dfrac{(3n)!\left(\frac{1}{27}\right)^n}{(n!)^3}$ converges or diverges. Any help?
 A: Observe that
\begin{align}
\sum_{n=1}^\infty\frac{(3n)!\left(\frac{1}{27}\right)^n}{(n!)^3}
&=\sum_{n=1}^\infty\frac{(3n)!}{\left[3^n(n!)\right]^3}\\
&=\sum_{n=1}^\infty\frac{(3n)!}{\left[3\cdot6\cdot9\cdots3n\right]^3}\\
&=\sum_{n=1}^\infty\left[\frac{1\cdot2}{3^2}\cdot\frac{4\cdot5}{6^2}\cdots\frac{(3n-2)(3n-1)}{(3n)^2}\right].
\end{align}
By using the limit comparison test with the divergent 
series $\displaystyle\sum_{n=1}^\infty\frac{1}{n}$, we see that
\begin{align}
&\lim_{n\to\infty}\frac{\frac{1\cdot2}{3^2}\cdot\frac{4\cdot5}{6^2}\cdots\frac{(3n-2)(3n-1)}{(3n)^2}}{\frac{1}{n}}\\
&\qquad=\lim_{n\to\infty}\left[\frac{1\cdot2}{3^2}\cdot\frac{4\cdot5}{6^2}\cdots
\frac{(3n-5)(3n-4)}{(3n-3)^2}\cdot\frac{(3n-2)(3n-1)}{9n}\right]\\
&\qquad=\lim_{n\to\infty}\left[\frac{1\cdot2}{3^2}\cdot\frac{4\cdot5}{3\cdot6}\cdots
\frac{(3n-5)(3n-4)}{(3n-6)(3n-3)}\cdot\frac{(3n-2)(3n-1)}{(3n-3)3n}\right]\\
&\qquad\ge \frac{1\cdot2}{3^2}=\frac{2}{9}.
\end{align}
Hence $\displaystyle\sum_{n=1}^\infty\frac{(3n)!\left(\frac{1}{27}\right)^n}{(n!)^3}$ diverges.
A: Using Stirling approximation we obtain
$$
\binom{3n}{n,n,n} \sim c \frac{3^{3n}}{n}
$$
for some $c>0$. Then
$$
\sum_{n\ge 1}\frac{1}{3^{3n}}\binom{3n}{n,n,n} \sim c\sum_{n\ge 1}\frac{1}{n} \sim c\ln n,
$$
which diverges.
