Help on how to prove that $\ln{k\choose k/2}=\sum_{n=1}^{\infty}(-1)^{n-1}\left({k\over n}-\ln{n+k\over n}\right)$ 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 $(5)?$
 A: 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.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\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}
