$\sum_{k\geq 1}\left(1-2k\,\text{arctanh}\frac{1}{2k}\right)=\frac{\log 2-1}{2}$ - looking for an elementary solution
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\sum_{k\ \geq\ 1}\bracks{1 - 2k\,\mrm{arctanh}\pars{1 \over 2k}} = {\ln\pars{2} - 1 \over 2}:\ ?}$.
$$\bbx{\ds{% \begin{array}{c} \mbox{Indeed, the partial sum can be evaluated explicitly by}\ elementary\ means\mbox{:} \\[3mm] \ds{\sum_{k = 1}^{N}\bracks{1 - 2k\,\mrm{arctanh}\pars{1 \over 2k}} = N - N\ln\pars{2N + 1} - N\ln\pars{2} + \ln\pars{\bracks{2N}! \over N!}} \\[3mm] = \ds{N\bracks{1 - \ln\pars{1 + {1 \over 2N}}} - 2N\ln\pars{2} -N\ln\pars{N} + \ln\pars{\bracks{2N}! \over N!}} \\[3mm] \mbox{The proof is at the}\ \ds{\color{#f00}{very\ end}}. \end{array}}} $$ It yields the correct limit: $$ \lim_{N \to \infty}\bracks{N - N\ln\pars{2N + 1} - N\ln\pars{2} + \ln\pars{\bracks{2N}! \over N!}} = \bbox[#ffe,10px,border:1px dotted navy]{\ds{% {\ln\pars{2} - 1 \over 2}}} $$ because $$ \ln\pars{\bracks{2N}! \over N!} \,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, \pars{2N + \color{#f00}{1 \over 2}}\ln\pars{2} + N\ln\pars{N} - N $$ $\ds{\color{#f00}{Without\ Stirling}}$, it's difficult to recover the crucial above $\ds{\color{#f00}{red}}$ mentioned $\ds{\color{#f00}{1 \over 2}}$ factor. Namely, \begin{align} \ln\pars{\bracks{2N}! \over N!} & = \sum_{k = 1}^{N}\ln\pars{k + N} = N\ln\pars{N} + N\ \overbrace{\bracks{{1 \over N}\sum_{k = 1}^{N}\ln\pars{1 + {k \over N}}}} ^{\ds{\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,2\ln\pars{2} - 1}} \\[5mm] & \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, \pars{2N + \color{#f00}{0}}\ln\pars{2} + N\ln\pars{N} - N \end{align}
Finite Sum: \begin{align} &\sum_{k = 1}^{N}\bracks{1 - 2k\,\mrm{arctanh}\pars{1 \over 2k}} = \sum_{k = 1}^{N}\bracks{1 - k\ln\pars{2k + 1 \over 2k - 1}} \\[5mm] = &\ N - \sum_{k = 1}^{N}k\ln\pars{2k + 1} + \sum_{k = 1}^{N}k\ln\pars{2k - 1} \\[5mm] = &\ N - \sum_{k = 1}^{N}k\ln\pars{2k + 1} + \sum_{k = 1}^{N - 1}\pars{k + 1}\ln\pars{2k + 1} \\[5mm] = &\ N - \sum_{k = 1}^{N}k\ln\pars{2k + 1} + \sum_{k = 1}^{N}\pars{k + 1}\ln\pars{2k + 1} - \pars{N + 1}\ln\pars{2N + 1} \\[5mm] = &\ N - \pars{N + 1}\ln\pars{2N + 1} + \sum_{k = 1}^{N}\ln\pars{2k + 1} \\[5mm] = &\ N - \pars{N + 1}\ln\pars{2N + 1} + \sum_{k = 3}^{2N + 1}\ln\pars{k} - \sum_{k = 2}^{N}\ln\pars{2k} \\[1cm] = &\ N - \pars{N + 1}\ln\pars{2N + 1} + \bracks{-\ln\pars{2} + \sum_{k = 2}^{2N}\ln\pars{k} + \ln\pars{2N + 1}} \\[5mm] - &\ \bracks{\pars{N - 1}\ln\pars{2} + \sum_{k = 2}^{N}\ln\pars{k}} \\[1cm] = &\ \bbx{N - N\ln\pars{2N + 1} - N\ln\pars{2} + \ln\pars{\bracks{2N}! \over N!}} \end{align}
Write $$1 - 2 k \arctan\left(\frac{1}{2k}\right) = \int_0^1 \dfrac{x^2}{x^2-4k^2}\; dx$$ Now, although you said you wanted to avoid it, the sum
$$ \sum_{k=1}^\infty \dfrac{x^2}{x^2-4k^2} = \dfrac{\pi x}{4} \cot\left(\frac{\pi x}{2}\right) - \frac{1}{2}$$
isn't so bad: using partial fractions, $$ \sum_{k=1}^N \dfrac{x^2}{x^2-4k^2} = \frac{x}{4} \left( \Psi(N+1+x/2) - \Psi(N+1-x/2) + \Psi(1-x/2) - \Psi(1+x/2)\right) $$ As $N \to \infty$, $\Psi(N+1+x/2) - \Psi(N+1-x/2) \to 0$ because $\lim_{t \to + \infty} \Psi'(t) = 0$, so that we are left with $$\sum_{k=1}^\infty \dfrac{x^2}{x^2-4k^2} = \frac{x}{4} \left(\Psi(1-x/2) - \Psi(1+x/2)\right)$$ which by the reflection formula is $\frac{\pi x}{4} \cot \left(\frac{\pi x}{2}\right) - \frac{1}{2}$. Then integrate this from $x=0$ to $1$ to get $ \dfrac{\ln 2}{2} - \dfrac{1}{2}$.