Evaluating the ratio $ {{a_{n+1}}\over{a_n}}$ in calculating the radius of convergence for a power series In calculating the radius of convergence for the power series
$$ \sum_{n=1}^\infty {{(2n)!}\over(n!)^2}\ x^n $$
By the ratio test, we let
$$ a_n = \lvert {{(2n)!}\over(n!)^2}\ x^n \rvert \quad\quad a_{n+1} = \lvert{{(2n+2)!}\over(n+1)!^2}\ x^{n+1} \rvert$$
$$ {{a_{n+1}}\over{a_n}} = \lvert {(2n+2)!\over(2n)!}{(n!)^2\over(n+1)!^2}\ x \rvert \longrightarrow \boxed{4\lvert x\rvert < 1} \longrightarrow \boxed{\lvert x\rvert<\frac 14}$$
My method for deriving 4 as the limit of convergence was obtained crudely simply by inputting large values for n and calculating manually; I am unsure as to how to derive 4 formally by simplifying the ratio.
Similarly
$$ \sum_{n=1}^\infty {(n!)^3\over(3n)!}\ x^n$$
By the ratio test
$$ a_n=\lvert {(n!)^3\over(3n)!}\ x^n \rvert \quad\quad a_{n+1} = \lvert {(n+1)!^3\over(3n+3)!}\ x^{n+1} \rvert$$
$$ {{a_{n+1}}\over{a_n}} = \lvert {(n+1)!^3\over(n!)^3}{(3n)!\over(3n+3)!}\ x \rvert \longrightarrow \boxed{{1\over27}\lvert x\rvert<1} \longrightarrow \boxed{\lvert x\rvert<27}$$
Again, a formal method for deriving ${1\over27}$ would be much appreciated.

Hint 
\begin{equation}
\frac{(2n + 2)!}{(2n)!} \frac{(n!)^2}{(n + 1)!} = \frac{(2n + 2)(2n + 1)}{(n + 1)(n + 1)} = \frac{4n^2 + 6n + 2}{n^2 + 2n + 1} = \frac{4 + \frac{6}{n} + \frac{2}{n^2}}{1 + \frac{2}{n} + \frac{1}{n^2}}
\end{equation}
