TITLE: $\sum_{k\geq 1}\left(1-2k\,\text{arctanh}\frac{1}{2k}\right)=\frac{\log 2-1}{2}$ - looking for an elementary solution
QUESTION [6 upvotes]: As stated in the title, I am looking for the most elementary proof of the following identity:

$$ \sum_{k\geq 1}\left(1-2k\,\text{arctanh}\frac{1}{2k}\right) =
 \frac{\log 2-1}{2}\tag{1}$$

I have a proof that exploits $2\,\text{arctanh}\frac{1}{2k}= \log(2k+1)-\log(2k-1)$, summation by parts and Stirling's inequality, but I have the strong feeling I am missing something quite trivial, maybe related with some Riemann sum or with
$$ \sum_{k\geq 1}\left(1-2k\,\text{arctanh}\frac{1}{2k}\right) = -\sum_{m\geq 1}\frac{\zeta(2m)}{4^m(2m+1)}. \tag{2}$$
Any help is appreciated, thanks in advance.

I forgot to mention that I would like to avoid proving
$$\forall t\in(0,1),\qquad \sum_{k\geq 1}\frac{4t^2}{4k^2-t^2}=2-\pi t \cot\frac{\pi t}{2} \tag{3}$$
for first. That clearly allows us to compute the LHS of $(1)$ as an integral, but requires Herglotz trick or something similar (Weierstrass products, digamma function, whatever).

REPLY [6 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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$\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}