I am trying to evaluate $$\lim_{n\to \infty} \bigg(\frac{(2n!)}{n!^2}\bigg)^{\frac{1}{4n}},$$ which came from trying to find the radius of convergence of the complex power series $$\sum_{n\ge0} z^{2n}\frac{\sqrt{(2n)!}}{n!}$$ In the limit I we have some cancellation: $$\frac{(2n!)}{n!^2}=\frac{1\cdots n\cdot (n+1)\cdots(2n)}{1\cdots n\cdot 1\cdots n}=\frac{(n+1)\cdot(2n)}{1\cdots n},\,\,\,(*)$$ and the right hand side is less than $2^n$, so an upper bound on the limit is $$(2^n)^{\frac{1}{4n}}=2^{1/4}$$ The right hand side of $(*)$ is at least $1$, so I have the bounds $1,2^{1/4}$, but I don't know how to get a better estimate. How can I compute this limit?
$$\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)$$  Now, $\log(n!) = \Theta (n\log(n))$ so I think we could write,   $$\lim_{n\to\infty}\frac{1}{2n}\left(\log\left(2n!\right) - \log\left(n!^2\right)\right) = \frac{1}{2n}\left(\log\left(2n!\right) - 2\log\left(n!\right)\right) $$  $$\lim_{n\to\infty}\frac{1}{2n}\left(2n\log\left(2n\right) - 2n\log\left(n\right)\right) $$  $$\log\left(2n\right) - \log\left(n\right) = \log(2)$$  Is this legitimate? I feel this might be wrong since, the $\Theta$ notation conceals a constant factor.
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