How would you show that the series $\sum_{n=1}^\infty \frac{(2n)!}{4^n (n!)^2}$ diverges? How would you show that the series $$\sum_{n=1}^\infty \frac{(2n)!}{4^n (n!)^2}$$ diverges? Wolfram Alpha says it diverges "by comparison", but I'd like to know to what you would compare it? I've tried some basic things to no avail: The ratio test is inconclusive, comparing with something "smaller" which I know diverges is proving difficult. If possible, I'd like to show this without using anything fancy (such as Stirling's approximation).
Can you do this with a "basic" comparison?

After the great answer of Adam Hughes one likely needs no more...        
However, I like the following comparision-test, at which I arrived after some fiddling.       
We compare $\sum a_n $ with $\sum \frac1{2n}$ and show by induction, that always $a_n>\frac1{2n}$  and the series diverges by comparision with the harmonic series.
Let's denote the series 
$$ s=\sum_{k=0}^\infty a_n  = 1+  \frac12+\frac12\frac34+\frac12\frac34\frac56 + \frac12\frac34\frac56\frac78 + ...\\
$$
Rewritten in the spirit of the said comparision this is
$$ s =1+  \frac12\left[(1)1 +\left({3\over2}\right)\frac12+\left({ 3 \cdot 5\over 2\cdot 4}\right)\frac13+\left({ 3 \cdot 5 \cdot 7\over 2\cdot 4 \cdot 6}\right)\frac14 + ... \right] \\
$$
and we want to show, that all parentheses have value greater than 1.
Induction as follows: we observe that this is true for the parenthese $p_2= \left( \frac32\right)$ at $a_2$: $\left({ 3\over2}\right) \gt 1$
The next parenthese is the previous postmultiplied by $\frac54$ so also this must be greater than 1. And this is obviously the case for all following parentheses.        
Formally:       


*

*Let  $\displaystyle \qquad a_n = {1 \cdot 3 \cdot ... \cdot (2n-1)\over 2^n n!} $
and $\displaystyle \qquad p_n = {1 \cdot 3 \cdot ... \cdot (2n-1)\over 2^{n-1} (n-1)!} $        

*Then  $ \displaystyle \qquad p_n\gt1 \implies p_{n+1}=p_n \cdot {2n+1\over2n} \gt 1$          

*and because we know that $ \displaystyle \qquad p_2>1 $           


all parentheses $p_{n \gt 1} \gt 1$ and all terms $a_{n \gt 1} \gt \frac1{2n}$ and the series diverges.

After @Cameron Williams has shown, that even $a_n \gt {1\over \sqrt{\pi} n} \approx {1\over 1.7724} \frac1n $ I used regression to find an even tighter estimate for the $a_n$ in relation to the harmonic numbers. If I did not have a stupid bug with this I found something like $$ p_n ^2 \gt -0.317894293187 +1.27323781105n \approx {4n-1\over \pi} \\ \implies a_n \gt \sqrt{\small  {4n-1\over \pi }} \cdot \frac1n$$
(The estimate ${4n-1\over \pi}$ is due to the comments of @Claude Leibovici)
