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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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
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}
Felix Marin
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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 $\sum_{k=1}^n \frac{n}{n^2+k^2}$
lab bhattacharjee
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how do you jump from the 3rd line to the blue line? – The Substitute Nov 04 '14 at 17:36
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@TheSubstitute, Have you noticed the last line of the other answer ? $$\text{as }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$ – lab bhattacharjee Nov 04 '14 at 17:40