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How can we show that $(1)$

$$\ln{k\choose k/2}=\sum_{n=1}^{\infty}(-1)^{n-1}\left({k\over n}-\ln{n+k\over n}\right)\tag1$$

$$\ln{k\choose k/2}=k\ln{2}-\ln{\prod_{n=1}^{\infty}\left({n+k\over n}\right)^{(-1)^{n-1}}}\tag2$$

$$\ln{k! \over (k/2)!^2}=k\ln{2}-\ln{\prod_{n=1}^{\infty}\left({n+k\over n}\right)^{(-1)^{n-1}}}\tag3$$

$$-\ln{k! \over (k/2)!^2}\cdot{1\over 2^k}=\ln{\prod_{n=1}^{\infty}\left({n+k\over n}\right)^{(-1)^{n-1}}}\tag4$$

$${k! \over (k/2)!^2}\cdot{1\over 2^k}={\prod_{n=1}^{\infty}\left({n+k\over n}\right)^{(-1)^{n}}}\tag5$$

How do I continue from $(5)?$

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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mrm}[1]{\,\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\ln\left(k \atop k/2\right) = \sum_{n = 1}^{\infty}\pars{-1}^{\, n - 1}\, \bracks{{k\over n} - \ln\pars{n + k \over n}}:\ ?}$


\begin{align} &\color{#f00}{\sum_{n = 1}^{\infty}\pars{-1}^{\, n - 1}\, \bracks{{k\over n} - \ln\pars{n + k \over n}}} = k\ln\pars{2} + \sum_{n = 1}^{\infty}\pars{-1}^{\, n}\,\ln\pars{1 + {k \over n}} \\[5mm] & =\ k\ln\pars{2} - \sum_{n = 0}^{\infty}\pars{-1}^{\, n}\int_{0}^{1}{k \over n + 1 + kt}\,\dd t = k\ln\pars{2} - k\int_{0}^{1}\sum_{n = 0}^{\infty}{\pars{-1}^{\, n} \over n + 1 + kt}\,\dd t \\[5mm] = &\ k\ln\pars{2} + k\int_{0}^{1}\sum_{n = 0}^{\infty} \pars{{1 \over 2n + 2 + kt} - {1 \over 2n + 1 + kt}}\,\dd t \\[5mm] & =\ k\ln\pars{2} + \half\,k\int_{0}^{1}\bracks{% \Psi\pars{\half\,kt + \half} - \Psi\pars{\half\,kt + 1}}\,\dd t\quad \pars{~\Psi:\ Digamma\ Function~} \end{align}
\begin{align} &\color{#f00}{\sum_{n = 1}^{\infty}\pars{-1}^{\, n - 1}\, \bracks{{k\over n} - \ln\pars{n + k \over n}}} \\[5mm] = &\ k\ln\pars{2} + \half\,k\,\,\bracks{\vphantom{\huge A^{A}}{2 \over k}% \ln\pars{\Gamma\pars{kt/2 + 1/2} \over \Gamma\pars{kt/2 + 1}}} _{\ t\ =\ 0}^{\ t\ =\ 1}\qquad \pars{~\Gamma:\ Gamma\ Function~} \\[5mm] = &\ k\ln\pars{2} + \ln\pars{{\Gamma\pars{k/2 + 1/2} \over \Gamma\pars{k/2 + 1}}\, {\Gamma\pars{1} \over \Gamma\pars{1/2}}} \\[5mm] = &\ \ln\pars{{2^{k} \over \pars{k/2}!}\, {\Gamma\pars{k/2 + 1/2} \over \root{\pi}}}\,,\qquad\qquad \pars{~\Gamma\pars{1} = 1\,,\ \Gamma\pars{\half} = \root{\pi}~}\tag{1} \end{align} However, with the $\ds{\Gamma}$-Duplication Formula \begin{align} \Gamma\pars{{k \over 2} + \half} & = {\root{2\pi}2^{1/2 - k}\,\,\,\Gamma\pars{k} \over \Gamma\pars{k/2}} = 2\root{\pi}2^{-k}\,{\pars{k - 1}! \over \pars{k/2 - 1}!} = 2\root{\pi}2^{-k}\,\,{k! \over k}\,{k/2 \over \pars{k/2}!} \\[5mm] & = \root{\pi}2^{-k}\,\,{k! \over \pars{k/2}!} \end{align} With this result, $\ds{\pars{1}}$ is reduced to: \begin{align} \color{#f00}{\sum_{n = 1}^{\infty}\pars{-1}^{\, n - 1}\, \bracks{{k\over n} - \ln\pars{n + k \over n}}} & = \ln\pars{k! \over \pars{k/2}!\pars{k/2}!} = \color{#f00}{\ln\left(\, k \atop k/2\,\right)} \end{align}

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We have $$\prod_{n=1}^{2N}\left(\frac{n+k}{n}\right)^{\left(-1\right)^{n}}=\prod_{n=1}^{N}\frac{\left(2n-1\right)\left(2n+k\right)}{2n\left(2n-1+k\right)}=\frac{\left(2N-1\right)!!\left(\frac{k}{2}+1\right)_{N}}{\left(2N\right)!!\left(\frac{k+1}{2}\right)_{N}}$$ where $\left(x\right)_{k}$ is the Pochhammer' symbol. Using the well known identities of the double factorial, we have $$\prod_{n=1}^{2N}\left(\frac{n+k}{n}\right)^{\left(-1\right)^{n}}=\frac{\Gamma\left(N+\frac{1}{2}\right)\left(\frac{k}{2}+1\right)_{N}}{\sqrt{\pi}N!\left(\frac{k+1}{2}\right)_{N}}$$ and now using the Stirling's approximation for the Gamma function and for the Pochhammer' symbol we have $$\prod_{n=1}^{\infty}\left(\frac{n+k}{n}\right)^{\left(-1\right)^{n}}=\lim_{N\rightarrow\infty}\frac{\Gamma\left(N+\frac{1}{2}\right)\left(\frac{k}{2}+1\right)_{N}}{\sqrt{\pi}N!\left(\frac{k+1}{2}\right)_{N}}=\color{red}{\frac{1}{\sqrt{\pi}}\frac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k}{2}+1\right)}}$$ which is equivalent to your claim.

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