Radius of convergence of $\displaystyle{\sum_{n=0}^{\infty}} {(n!)^3 \over (3n)!}z^{3n} $. 
Find the radius of convergence of $\displaystyle{\sum_{n=0}^{\infty}} {(n!)^3 \over (3n)!}z^{3n} \ ?$

I applied Cauchy-Hadamard test and the result is coming $0$ (radius of convergence). To obtain the limit I also used Cauchy's first limit theorem. For a lot of messy calculation I didn't provide my work.
Please someone check whether I'm right or wrong.
Thank you..
 A: By the ratio test the sum converges if $\lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n}  = r$ and $|r| <1$ , special cases are when $|r|=1$ and have to be checked separately.
$a_n = \frac{(n !)^3}{(3n)!} z^{3n}$ so $\lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n} =\frac{((n+1)!)^3 z^{3 n+3}}{\frac{(3 (n+1))! \left((n!)^3 z^{3 n}\right)}{(3
   n)!}} = \frac{z^3 (3 n)! ((1+n)!)^3}{(n!)^3 (3 (1+n))!} = \frac{(1+n)^2 z^3}{3 (2+9 n (1+n))}=|z^3| \lim \limits_{n \to \infty}{\frac{(1+n)^2}{3(2+9n(1+n))}} = |z^3| \frac{1}{27} $
So for $|\frac{z^3}{27}| <1 $ its obvious that $-3 < z < 3$ (if you are looking for the domain of convergence for complex $z$, from here you have to continue dealing with $z$ as complex number, the methods above works for real and complex).
So when $|\frac{z^3}{27}|=1$ we get that $z=-3,3$, we need to check them.
Just substitute instead of $z$ the numbers $-3$ and then $3$ in the sum, in both cases the sum diverges using other tests than the ratio test because its inconclusive when the ratio is exactly one.
A: Hint:
First find the radius of convergence of the series $\;\displaystyle\sum_{n=0}^{\infty} {(n!)^3 \over (3n)!}z^{n}$.
For this apply Cauchy-Hadamard's formula, combined with Stirling's formula:
\begin{align}
\biggl(\frac{(n!)^3}{(3n)!}\biggr)^{\!\tfrac1n}\sim_\infty&\frac{\Bigl(\sqrt{(2\pi n)^{3\mathstrut}}\Bigr)^{\!\tfrac1n}\Bigl(\dfrac n{\mathrm e}\Bigr)^{\!3}}{\Bigl(\sqrt{6\pi n\mathstrut}\Bigr)^{\!\tfrac1n}\Bigl(\dfrac{3n}{\mathrm e}\Bigr)^{\!3}}=\frac{1}{27}\sqrt{\Bigl(\frac 43\pi^2n^2\Bigr)^{_\tfrac1n}},
\end{align}
and we obtain a radius of convergence equal to $27$ for the auxiliary series.
A: If we represent the factorials everywhere as gamma-functions and Pochhammer symbols, this is actually a generalized hypergeometric function 4F3(...;;;z/3). According to wikipedia the radius of convergence is where the argument becomes 1, which means in our case, |z|=3.
