Evaluating $\lim_{n\to \infty} \bigg(\frac{(2n!)}{n!^2}\bigg)^{\frac{1}{4n}}$ 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?
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With Stirling Asymptotic formula:

\begin{align}&\color{#66f}{\large\lim_{n\ \to\ \infty}
\bracks{\pars{2n}! \over \pars{n!}^{2}}^{1/\pars{4n}}}
=\lim_{n\ \to\ \infty}
\bracks{\root{2\pi}\pars{2n}^{2n + 1/2}\expo{-2n}\over
\pars{\root{2\pi}n^{n + 1/2}\expo{-n}}^{2}}^{1/\pars{4n}}
\\[5mm]&=\lim_{n\ \to\ \infty}\bracks{%
\pars{2\pi}^{-1/2}2^{2n + 1/2}n^{n}}^{1/\pars{4n}}
=\lim_{n\ \to\ \infty}\bracks{\pars{2\pi}^{-1/\pars{8n}}\ 
2^{1/2 + 1/\pars{8n}}\ n^{-1/\pars{8n}}}
=2^{1/2}
\\[5mm]&= \color{#66f}{\Large\root{2}} \approx {\tt 1.4142}\quad\mbox{since}\quad
\left\{\begin{array}{rcl}
\lim_{n\ \to\ \infty}\pars{2\pi}^{-1/\pars{8n}} & = & 1
\\[2mm]
\lim_{n\ \to\ \infty}2^{1/2 + 1/\pars{8n}} & = & 2^{1/2} = \root{2}
\\[2mm]
\lim_{n\ \to\ \infty}n^{-1/\pars{8n}} & = & 1
\end{array}\right.
\end{align}
A: Let $$A=\lim_{n\to \infty} \bigg(\frac{(2n!)}{n!^2}\bigg)^{\frac{1}{4n}}$$
$$\implies\ln A=\lim_{n\to \infty}\frac1{4n}\sum_{r=1}^n\ln\frac{n+r}r$$
$$\implies4\ln A=\lim_{n\to \infty}\frac1n\sum_{r=1}^n\ln\frac{1+r/n}{r/n}$$
the rest should be like The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$
