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?

Let $$(2n)!!=\displaystyle\prod_{\substack{2\le x\le 2n\\ x\equiv 0 \pmod 2}} x$$and $$(2n)!!^\ast=\displaystyle\prod_{\substack{2\le x\le 2n\\ x\equiv 1 \pmod 2}} x\tag{1}$$  Then, $$(2n)!!=2^n n!$$ Hence, $$\dfrac{(2n)!}{4^n(n!)^2}=\dfrac{(2n)!}{((2n)!!)^2}=\dfrac{(2n)!!^\ast}{(2n)!!}\ge \left(\dfrac{2n-1}{2n-2}\right)^{n-1}\cdot\dfrac{1}{2n}\ge \dfrac{1}{2n}\tag{2}$$Consequently, $$\displaystyle\sum_{n=1}^\infty\dfrac{(2n)!}{4^n(n!)^2}\ge \displaystyle\sum_{n=1}^\infty\dfrac{1}{2n}$$But the series $\displaystyle\sum_{n=1}^\infty\dfrac{1}{2n}$ diverges and we are done. (Can you prove $(1)$ and $(2)$?)
Remarks. You may also use the method of this answer to show that $\displaystyle\sum_{n=1}^\infty\dfrac{(2n)!}{4^n(n!)^2}\ge \displaystyle\sum_{n=1}^\infty\dfrac{1}{2n}$.
