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

Question

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.

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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).

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{k\ \geq\ 1}\bracks{1 - 2k\,\mrm{arctanh}\pars{1 \over 2k}} = {\ln\pars{2} - 1 \over 2}:\ ?}$.

$$\bbx{\ds{% \begin{array}{c} \mbox{Indeed, the partial sum can be evaluated explicitly by}\ elementary\ means\mbox{:} \\[3mm] \ds{\sum_{k = 1}^{N}\bracks{1 - 2k\,\mrm{arctanh}\pars{1 \over 2k}} = N - N\ln\pars{2N + 1} - N\ln\pars{2} + \ln\pars{\bracks{2N}! \over N!}} \\[3mm] = \ds{N\bracks{1 - \ln\pars{1 + {1 \over 2N}}} - 2N\ln\pars{2} -N\ln\pars{N} + \ln\pars{\bracks{2N}! \over N!}} \\[3mm] \mbox{The proof is at the}\ \ds{\color{#f00}{very\ end}}. \end{array}}} $$ 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}

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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}

Answer 2

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}$.

Answer 3

$$ \begin{align} & \zeta^{\ast}(2n)=\zeta(2n)-\frac{2}{2^{2n}}\zeta(2n) \Rightarrow \frac{\zeta(2n)}{2^{2n}}=\frac{\zeta(2n)-\zeta^{\ast}(2n)}{2} \Rightarrow \\[4mm] & \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2^{2n}(2n+1)} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{\zeta(2n)-\zeta^{\ast}(2n)}{2n+1} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{\Gamma(2n)\zeta(2n)-\Gamma(2n)\zeta^{\ast}(2n)}{\Gamma(2n)(2n+1)} = \\[4mm] & \frac{1}{2} \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} \int_{0}^{\infty} \left(\frac{x^{2n-1}}{e^x-1}-\frac{x^{2n-1}}{e^x+1}\right) \,dx = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} \int_{0}^{\infty} \frac{x^{2n-1}}{e^{2x}-1} \,dx = \\[4mm] & \int_{0}^{\infty} \frac{1}{e^{2x}-1} \sum_{n=1}^{\infty} \frac{(2n\color{red}{+1-1}) \space x^{2n-1}}{(2n+1)!} \,dx = \int_{0}^{\infty} \frac{1}{e^{2x}-1} \sum_{n=1}^{\infty} \left( \frac{x^{2n-1}}{(2n)!}-\frac{x^{2n-1}}{(2n+1)!} \right) \,dx = \\[4mm] & \int_{0}^{\infty} \left( \frac{x^{-1}}{e^{2x}-1} \sum_{n=1}^{\infty} \frac{x^{2n}}{(2n)!} - \frac{x^{-2}}{e^{2x}-1} \sum_{n=1}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \right) \,dx \normalsize = \\[4mm] & \int_{0}^{\infty} \left( \frac{\cosh(x)-1}{x (e^{2x}-1)} - \frac{\sinh(x)-x}{x^2 (e^{2x}-1)}\right) \,dx \\[4mm] & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \\ \end{align} $$ 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}

Answer 4

I didn't intend to actually answer this question, because I don't know the answer. So please consider this as thinking out loud. Here is what popped into my mind as an amateur. Let $$\tanh y=\frac1{2k}$$ then: $$\tanh y=\frac{e^y-e^{-y}}{e^y+e^{-y}}=\frac{e^{2y}-1}{e^{2y}+1}=\frac 1{2k}\implies e^{2y}=\frac{2k+1}{2k-1}$$ Therefore from $$y=\text{arctanh}\frac 1{2k}=\frac 1 2\ln\left(\frac{2k+1}{2k-1}\right)$$ one can write: $$\begin{align}\sum_{k=1}^\infty \left(1-2k\text{ arctanh}\frac 1{2k}\right)&=\sum_{k=1}^\infty(1-k\ln(2k+1)+k\ln(2k-1))\\ &=\ln\prod_{k=1}^\infty\left(\left( \frac{2k-1}{2k+1}\right)^ke\right) \end{align}$$ Then I messed around a bit with this: $$\begin{align}e^n\prod_{k=1}^n\left(\frac{2k-1}{2k+1}\right)^k&=e^n\left( \frac{1}{3}\right)^1\left( \frac{3}{5}\right)^2\cdots\left( \frac{2n-1}{2n+1}\right)^n\\ &=e^n\frac{1\times 3\times\cdots\times(2n-1)}{(2n+1)^n}\\ &=1\times 3\times\cdots\times(2n-1)\left(\frac{e}{2n+1}\right)^n\\ &=\frac{(2n)!}{n!\,2^n}\left(\frac{e}{2n+1}\right)^n=\frac{(2n)!}{n!\,4^n}\left(\frac{e}{n+1/2}\right)^n \end{align}$$ whose limit should eventually be equal to $\sqrt{2/e}$. But I don't know how to get there.