Convergence or divergence of $ \sum \frac{(2n)!}{2^{2n}(n!)^2}$ using Raabe's Test 
Determine convergence or divergence
  $$ \sum \frac{(2n)!}{2^{2n}(n!)^2}$$

Considering $a_n = \frac{(2n)!}{2^{2n}(n!)^2}$ for every $n$ s. t. $n \in [0, \infty)$
$$ \frac{a_{n+1}}{a_n}= \frac{(2(n+1))!}{2^{2(n+1)}((n+1)!)^2} \frac{2^{2n}(n!)^2}{(2n)!}.$$
After simplification,
$$ \frac{a_{n+1}}{a_n}= \frac{2n+1}{2n+2}.$$
Taking the limit:
$$\lim\limits_{n \rightarrow \infty} \frac{2+\frac{1}{n}}{2+\frac{2}{n}} =1.$$
As $L = 1$, the test is inconclusive.
Now, attempting Raabe's Test:


Raabe's test: suppose that $a_n>0$ for large $n$, let $$M =\overline\lim\limits_{n \rightarrow \infty} n \left( \frac{a_{n+1}}{a_n}-1 \right),\text { and } m = \underline \lim\limits_{n \rightarrow \infty} n \left( \frac{a_{n+1}}{a_n}-1 \right) . $$
    Then 
(a) $\sum a_n < \infty$, if $M< -1$
(b) $\sum a_n = \infty $, if $m > -1$
(c) the test is inconclusive if $m \leq -1 \leq M.$


Applying 
$$\lim\limits_{n \rightarrow \infty} n \left( \frac{a_{n+1}}{a_n}-1 \right)= \lim\limits_{n \rightarrow \infty} n\left(  \frac{2n+1}{2n+2}     -1 \right) = \lim\limits_{n \rightarrow \infty}   \frac{3n}{2n^2+2n} = 0> -1$$
(using L'Hospital).
Therefore $\sum a_n < \infty.$
Is my test correct? If yes, I do not understand how $\overline\lim$ and $\underline \lim$ is applied here and their meaning. How do you deal with $m$ and $M$? Is there a more efficient test for this?
Much appreciated
 A: If the limit exists then $m=M$ and they are equal to the limit. This is because there is only one sub sequential limit, so the infimum and supremum of the set of subsequential limits coincide.
Note also that $n\left(\frac{2n+1}{2n+2}-1 \right) = \frac{-n}{2n+2}$. So your limit should be $-1/2$.
A: This does not use any test to show divergence.
Consider $$a_n=\frac{(2n)!}{2^{2n}(n!)^2}$$ Take logarithms $$\log(a_n)=\log((2n)!)-2n\log(2)-2\log(n!)$$ and use Stirling approximation for large values of $p$ $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
   \left({p}\right)\right)+O\left({\frac{1}{p}}\right)$$ which makes $$\log(a_n)=-\frac{1}{2} \left(\log \left({n}\right)+\log (\pi
   )\right)+O\left(\frac{1}{n}\right)$$ Continuing with Taylor $$a_n=e^{\log(a_n)}=\frac 1{\sqrt{\pi n}}+O\left(\frac{1}{n^{3/2}}\right)$$ By comparison with $p$-series, then $\sum a_n$  diverges.
You also could consider $$y=\sum_{n=0}^\infty \frac{(2n)!}{2^{2n}(n!)^2} x^n$$ and recognize that this is the infinite Taylor expansion of $y=\frac{1}{\sqrt{1-x}}$ and think about what is going on when $x\to 1$.
Edit
You could notice that the problem comes from the $\color{red}{2}$. Considering
$$b_n=\frac{(2n)!}{k^{2n}(n!)^2}$$ we should get $$\frac{b_{n+1}}{b_n}=\frac{2(2n+1)}{k^2 (n+1)}$$ which is $<1$ as soon as $k >2$.
Eventually, you could recognize that
$$y=\sum_{n=0}^\infty \frac{(2n)!}{k^{2n}(n!)^2} x^n=\frac{k}{\sqrt{k^2-4 x}}$$ making $$\sum_{n=0}^\infty \frac{(2n)!}{k^{2n}(n!)^2}=\frac{k}{\sqrt{k^2-4 }}$$
