$\displaystyle \sum_{n=1}^{\infty}\frac{(n!)^2}{2n!}$

Unlike some other that can do this one in there heads, I have to workout it out visually, I used the ration test, I believe I may have made some mistakes in simplification so I am asking for a check like always.

$\displaystyle a_n=\frac{(n!)^2}{(2n!)}$

$\displaystyle a_n+1=\frac{(n+1!)^2}{(2n+2!)}$

$\displaystyle \lim_{n\to\infty}\|\frac{(n+1!)^2}{(2n+2!)}\cdot\f rac{(2n!)}{(n!)^2}$

Doing some simplification to

$\displaystyle \lim_{n\to\infty}\frac{(n+1)^2}{(2n+2)(2n+1)}$

Multiplying out the terms in the numerator and denominator gave me a fraction with the same degree in both denominator and numerator

$\displaystyle \lim_{n\to\infty}\frac{(n^2+2n+1)}{(4n^2+6n+2)}$

I can take the ratio of the degrees of the highest powers since they the same to find limit which gives $\displaystyle L = \frac{1}{4}$ and the ratio test states, if the limit is less than 1, the series converges, is this correct?

I just got confused when wolfram, simplified the series to $\displaystyle \sum_{n=1}^{\infty}\frac{n!}{2}$ and stated it diverges, I was using wolfram for a quick check, If I did things correct or not.