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