$\sum_{k\geq 1}\left(1-2k\,\text{arctanh}\frac{1}{2k}\right)=\frac{\log 2-1}{2}$ - looking for an elementary solution 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).

$$
\begin{align}
& \zeta^{\ast}(2n)=\zeta(2n)-\frac{2}{2^{2n}}\zeta(2n) \Rightarrow \frac{\zeta(2n)}{2^{2n}}=\frac{\zeta(2n)-\zeta^{\ast}(2n)}{2} \Rightarrow \\[4mm]
& \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2^{2n}(2n+1)} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{\zeta(2n)-\zeta^{\ast}(2n)}{2n+1} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{\Gamma(2n)\zeta(2n)-\Gamma(2n)\zeta^{\ast}(2n)}{\Gamma(2n)(2n+1)} = \\[4mm]
& \frac{1}{2} \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} \int_{0}^{\infty} \left(\frac{x^{2n-1}}{e^x-1}-\frac{x^{2n-1}}{e^x+1}\right) \,dx = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} \int_{0}^{\infty} \frac{x^{2n-1}}{e^{2x}-1} \,dx = \\[4mm]
& \int_{0}^{\infty} \frac{1}{e^{2x}-1} \sum_{n=1}^{\infty} \frac{(2n\color{red}{+1-1}) \space x^{2n-1}}{(2n+1)!} \,dx = \int_{0}^{\infty} \frac{1}{e^{2x}-1} \sum_{n=1}^{\infty} \left( \frac{x^{2n-1}}{(2n)!}-\frac{x^{2n-1}}{(2n+1)!} \right) \,dx = \\[4mm]
& \int_{0}^{\infty} \left( \frac{x^{-1}}{e^{2x}-1} \sum_{n=1}^{\infty} \frac{x^{2n}}{(2n)!} - \frac{x^{-2}}{e^{2x}-1} \sum_{n=1}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \right) \,dx \normalsize = \\[4mm]
& \int_{0}^{\infty} \left( \frac{\cosh(x)-1}{x (e^{2x}-1)} - \frac{\sinh(x)-x}{x^2 (e^{2x}-1)}\right) \,dx \\[4mm]
& \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \\
\end{align}
$$
Both integral are convergent, directly calculate:
$$
\int_{0}^{\infty} \frac{\cosh(x)-1}{x^1 (e^{2x}-1)} \,dx = \frac{1}{2}\left(\log{\pi}-\log{2}\right) \quad\&\quad \int_{0}^{\infty} \frac{\sinh(x)-x}{x^2 (e^{2x}-1)} \,dx = \frac{1}{2}\left(\log{\pi}-\log{e}\right) \\[4mm]
\Rightarrow \sum_{k=1}^{\infty} \left(1-2k\space\text{arctanh}\frac{1}{2k}\right) = - \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2^{2n}(2n+1)} = \frac{\log{2}-1}{2} \\[4mm]
$$
Or simplify first, and then calculate:
\begin{align}
& \int_{0}^{\infty} \left( \frac{\cosh(x)-1}{x (e^{2x}-1)} - \frac{\sinh(x)-x}{x^2 (e^{2x}-1)}\right) \,dx = \frac{1}{2} \int_{0}^{\infty} \frac{1}{x \space x} \left( \frac{x}{e^{x}+1} + \frac{x}{e^{x}-1} - \frac{x}{e^{x}} - \frac{1}{e^{x}} \right) \,dx \\[4mm]
& = \frac{1}{2} \int_{0}^{\infty} \frac{1}{x \space x} \left( \frac{x}{e^{x}+1} + \frac{x}{e^{x}-1} \color{red}{-1} - \frac{x}{e^{x}} - \frac{1}{e^{x}} \color{red}{+1} \right) \,dx \qquad \small \{\text{both convergent}\} \\[4mm]
& \int_{0}^{\infty} \frac{1}{x \space x} \left( \frac{x}{e^{x}+1} + \frac{x}{e^{x}-1} - 1 \right) \,dx = -\log{2} \\
& \{\small \lim_{x\rightarrow 0} [\Gamma(x)\zeta^{\ast}(x)+\Gamma(x)\zeta(x)] = \lim_{x\rightarrow 0} (2 - 2^{1-x}) \Gamma(x) \zeta(x) = -\log{2} \normalsize\} \\[4mm]
& \int_{0}^{\infty} \frac{1}{x \space x} \left( \frac{x}{e^{x}} + \frac{1}{e^{x}} - 1 \right) \,dx = -1 \\
& \{\small \lim_{x\rightarrow 0} [\Gamma(x) + \Gamma(x-1)] = \lim_{x\rightarrow 0} (1 - 1/(1-x)) \Gamma(x) = -1 \normalsize\} \\[4mm]
& \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \\
\end{align}
