General form for $\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$

Let

$$S\left( m \right)=\sum\limits_{n=1}^{\infty }{{{\left( -1 \right)}^{n}}\left( mn\operatorname{arcoth}\left( mn \right)-1 \right)},\,\,\left| m \right|>1$$

And note the logarithmic representation of the inverse function

$$\operatorname{arcoth}\left( z \right)=\frac{1}{2}\log \left( \frac{z+1}{z-1} \right)$$

The series can therefore be written as

$$S\left( m \right)=\sum\limits_{n=1}^{\infty }{\left( \log \left( \frac{1}{{{e}^{{{\left( -1 \right)}^{n}}}}}{{\left( \frac{nm+1}{nm-1} \right)}^{\frac{1}{2}{{\left( -1 \right)}^{n}}nm}} \right) \right)}=\log \left( \prod\limits_{n=1}^{\infty }{\frac{1}{{{e}^{{{\left( -1 \right)}^{n}}}}}{{\left( \frac{nm+{{\left( -1 \right)}^{n}}}{nm-{{\left( -1 \right)}^{n}}} \right)}^{\frac{1}{2}nm}}} \right)$$

Breaking the product into two convergent products consider then

$${{e}^{S}}=\prod\limits_{n=1}^{\infty }{\frac{1}{e}{{\left( \frac{2nm+1}{2nm-1} \right)}^{nm}}}\prod\limits_{n=1}^{\infty }{e{{\left( \frac{\left( 2n-1 \right)m-1}{\left( 2n-1 \right)m+1} \right)}^{\frac{1}{2}\left( 2n-1 \right)m}}}={{p}_{1}}\left( m \right){{p}_{2}}\left( m \right)$$

The first of these is

$${{p}_{1}}\left( m \right)=\prod\limits_{n=1}^{\infty }{\frac{1}{e}\frac{{{\left( 1+\frac{1}{2nm} \right)}^{nm}}}{{{\left( 1-\frac{1}{2nm} \right)}^{nm}}}}$$

However for the moment note that to ensure the convergence of

$${{p}_{a}}\left( m \right)=\prod\limits_{n=1}^{\infty }{{{\left( 1+\frac{1}{2nm} \right)}^{nm}}a}$$

we must consider the convergence of

$$\log {{p}_{a}}\left( m \right)=\sum\limits_{n=1}^{\infty }{nm\log \left( 1+\frac{1}{2nm} \right)+\log \left( a \right)}$$

For large n the summand behaves

$$nm\log \left( 1+\frac{1}{2nm} \right)+\log \left( a \right)\simeq \frac{1}{2}+\log \left( a \right)-\frac{1}{8mn}+O\left( {{n}^{-2}} \right)$$

so let $a={{e}^{\frac{1}{8mn}-\frac{1}{2}}}$. The product can now be written in the following manner

$${{p}_{1}}\left( m \right)=\prod\limits_{n=1}^{\infty }{\frac{{{\left( 1+\frac{1}{2nm} \right)}^{nm}}{{e}^{-\frac{1}{2}+\frac{1}{8mn}}}}{{{\left( 1-\frac{1}{2nm} \right)}^{nm}}{{e}^{\frac{1}{2}+\frac{1}{8mn}}}}}=\frac{\prod\limits_{n=1}^{\infty }{{{\left( 1+\frac{1}{2nm} \right)}^{nm}}{{e}^{-\frac{1}{2}+\frac{1}{8mn}}}}}{\prod\limits_{n=1}^{\infty }{{{\left( 1-\frac{1}{2nm} \right)}^{nm}}{{e}^{\frac{1}{2}+\frac{1}{8mn}}}}}$$

Therefore consider now

$${{p}_{1\pm }}=\prod\limits_{n=1}^{\infty }{{{\left( 1\pm \frac{1}{2nm} \right)}^{nm}}{{e}^{\mp \frac{1}{2}+\frac{1}{8mn}}}}$$

or alternatively

$$\log {{p}_{1\pm }}=\sum\limits_{n=1}^{\infty }{nm\log \left( 1\pm \frac{1}{2nm} \right)}\mp \frac{1}{2}+\frac{1}{8mn}$$

Using the series representation for the logarithm

$$\log {{p}_{1+}}=\sum\limits_{n=1}^{\infty }{nm\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k+3}}}{\left( k+2 \right){{\left( 2nm \right)}^{k+2}}}}}=-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\zeta \left( k+1 \right)}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}$$

$$\log {{p}_{1-}}=-\sum\limits_{n=1}^{\infty }{nm\sum\limits_{k=3}^{\infty }{\frac{1}{k{{\left( 2nm \right)}^{k}}}}}=-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{\zeta \left( k+1 \right)}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}$$

From Srivastava [pg. 257, (63)] the following series takes the form

$$\begin{align}\sum\limits_{k=2}^{\infty }{\frac{\zeta \left( k,a \right)}{k+n}{{x}^{k+n}}}&=\sum\limits_{k=0}^{n}{\left( \begin{matrix} n \\ k \\ \end{matrix} \right)\zeta '\left( -k,a-x \right){{x}^{n-k}}}-\sum\limits_{k=0}^{n-1}{\frac{\zeta \left( -k,a \right)}{n-k}{{x}^{n-k}}}\\&+\left\{ \psi \left( a \right)-{{H}_{n}} \right\}\frac{{{x}^{n+1}}}{n+1}-\zeta '\left( -n,a \right),\ \ \left| x \right|<\left| a \right|,\,\,n=0,1,2...\end{align}$$

where $\zeta '\left( s,a \right)=\frac{\partial }{\partial s}\zeta \left( s,a \right)$ . Hence

$$\sum\limits_{k=1}^{\infty }{\frac{\zeta \left( k+1 \right)}{k+2}{{x}^{k+1}}}=\frac{1}{x}\left\{ \zeta '\left( 0,1-x \right)x+\zeta '\left( -1,1-x \right)+\frac{1}{2}\left( x-{{x}^{2}} \right)-\tfrac{1}{2}\gamma {{x}^{2}}-\zeta '\left( -1 \right) \right\}$$

We have then

$$\log {{p}_{-}}\left( m \right)=-\frac{1}{2}\zeta '\left( 0,1-\frac{1}{2m} \right)-m\zeta '\left( -1,1-\frac{1}{2m} \right)-\frac{1}{4}\left( 1-\frac{\gamma +1}{2m} \right)+m\zeta '\left( -1 \right)$$

$$\log {{p}_{+}}\left( m \right)=\frac{1}{2}\zeta '\left( 0,1+\frac{1}{2m} \right)-m\zeta '\left( -1,1+\frac{1}{2m} \right)+\frac{1}{4}\left( 1+\frac{1+\gamma }{2m} \right)+m\zeta '\left( -1 \right)$$

Continuing, recall $${{p}_{2}}\left( m \right)=\prod\limits_{n=1}^{\infty }{e\frac{{{\left( 1-\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}}{{{\left( 1+\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}}}$$

and so for a moment consider the convergence of the product

$${{q}_{a}}=\prod\limits_{n=1}^{\infty }{{{\left( 1-\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}a}\Rightarrow a={{e}^{\frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}}$$

Therefore re-write the product

$$\begin{align}{{p}_{2}}\left( m \right)&=\prod\limits_{n=1}^{\infty }{\frac{{{\left( 1-\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}{{e}^{\frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}}}{{{\left( 1+\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}{{e}^{-\frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}}}}\\&=\frac{\prod\limits_{n=1}^{\infty }{{{\left( 1-\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}{{e}^{\frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}}}}{\prod\limits_{n=1}^{\infty }{{{\left( 1+\frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}{{e}^{-\frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}}}}\end{align}$$

Consider then the products and their log transforms

$${{p}_{2\pm }}=\prod\limits_{n=1}^{\infty }{{{\left( 1\pm \frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)}^{m\left( n-\tfrac{1}{2} \right)}}{{e}^{\mp \frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}}}$$ $$\log {{p}_{2\pm }}=\sum\limits_{n=1}^{\infty }{2m\left( n-\tfrac{1}{2} \right)\log \left( 1\pm \frac{1}{2m\left( n-\tfrac{1}{2} \right)} \right)\mp \frac{1}{2}+\frac{1}{8m\left( n-\tfrac{1}{2} \right)}}$$ Taking each case separately and expanding the logarithm in its series $$\log {{p}_{2+}}=-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}\sum\limits_{n=1}^{\infty }{\frac{1}{{{\left( n-\tfrac{1}{2} \right)}^{k+1}}}}=-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\left( -1+{{2}^{k+1}} \right)\zeta \left( k+1 \right)}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}$$ $$\log {{p}_{2-}}=-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{1}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}\sum\limits_{n=1}^{\infty }{\frac{1}{{{\left( n-\tfrac{1}{2} \right)}^{k+1}}}}=-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{\left( -1+{{2}^{k+1}} \right)\zeta \left( k+1 \right)}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}$$ So $$\log {{p}_{2+}}=\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\zeta \left( k+1 \right)}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\zeta \left( k+1 \right)}{\left( k+2 \right){{m}^{k+1}}}}=-\log {{p}_{1+}}\left( m \right)+\log {{p}_{1+}}\left( \tfrac{1}{2}m \right)$$ $$\log {{p}_{2-}}=\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{\zeta \left( k+1 \right)}{\left( k+2 \right){{\left( 2m \right)}^{k+1}}}}-\frac{1}{2}\sum\limits_{k=1}^{\infty }{\frac{\zeta \left( k+1 \right)}{\left( k+2 \right){{m}^{k+1}}}}=-\log {{p}_{1-}}\left( m \right)+\log {{p}_{1-}}\left( \tfrac{1}{2}m \right)$$ Therefore $${{p}_{2}}\left( m \right)=\frac{{{e}^{-\log {{p}_{1-}}\left( m \right)+\log {{p}_{1-}}\left( \tfrac{1}{2}m \right)}}}{{{e}^{-\log {{p}_{1+}}\left( m \right)+\log {{p}_{1+}}\left( \tfrac{1}{2}m \right)}}}$$ The final product is $${{e}^{S}}={{p}_{1}}\left( m \right){{p}_{2}}\left( m \right)={{e}^{2\log {{p}_{1+}}\left( m \right)-2\log {{p}_{1-}}\left( m \right)+\log {{p}_{1-}}\left( \tfrac{1}{2}m \right)-\log {{p}_{1+}}\left( \tfrac{1}{2}m \right)}}$$ and so the series becomes $$S=2\left\{ \log {{p}_{1+}}\left( m \right)-\log {{p}_{1-}}\left( m \right) \right\}+\log {{p}_{1-}}\left( \tfrac{1}{2}m \right)-\log {{p}_{1+}}\left( \tfrac{1}{2}m \right)$$

Substitution of previous results yields

$$\begin{align} S&=\zeta '\left( 0,1+\frac{1}{2m} \right)+\zeta '\left( 0,1-\frac{1}{2m} \right)-\frac{1}{2}\zeta '\left( 0,1-\frac{1}{m} \right)-\frac{1}{2}\zeta '\left( 0,1+\frac{1}{m} \right) \\ & -2m\zeta '\left( -1,1+\frac{1}{2m} \right)+2m\zeta '\left( -1,1-\frac{1}{2m} \right)-\frac{1}{2}m\zeta '\left( -1,1-\frac{1}{m} \right)\\&+\frac{1}{2}m\zeta '\left( -1,1+\frac{1}{m} \right)+\frac{1}{2} \\ \end{align}$$

Note: $\zeta '\left( 0,a \right)=\log \left( \Gamma \left( a \right) \right)-\tfrac{1}{2}\log \left( 2\pi \right)$, to obtain

$$\begin{align} S\left( m \right)&=\log \left( \frac{\Gamma \left( 1+\frac{1}{2m} \right)\Gamma \left( 1-\frac{1}{2m} \right)}{\sqrt{2\pi \Gamma \left( 1+\frac{1}{m} \right)\Gamma \left( 1-\frac{1}{m} \right)}} \right) \\ &+2m\left\{ \zeta '\left( -1,1-\frac{1}{2m} \right)-\zeta '\left( -1,1+\frac{1}{2m} \right) \right\}\\&+\frac{1}{2}m\left\{ \zeta '\left( -1,1+\frac{1}{m} \right)-\zeta '\left( -1,1-\frac{1}{m} \right) \right\}+\frac{1}{2} \\ \end{align}$$ Now fix m as a positive integer greater than one. Note also $\zeta \left( s,a+1 \right)=\zeta \left( s,a \right)-{{a}^{-s}}$ so $\zeta '\left( s,a+1 \right)=\zeta '\left( s,a \right)+{{a}^{-s}}\log \left( a \right)$ and DLMF 25.11.21

$$\begin{align} \zeta '\left( -1,\frac{h}{k} \right)&=\frac{\left( 1-\gamma -\ln \left( 2\pi k \right) \right)\left( \tfrac{1}{6}-\tfrac{h}{k}+{{\left( \tfrac{h}{k} \right)}^{2}} \right)}{2}-\frac{\left( 1-\gamma -\ln \left( 2\pi \right) \right)}{12{{k}^{2}}} \\ &+\frac{1}{4\pi {{k}^{2}}}\sum\limits_{r=1}^{k-1}{\sin \left( \frac{2\pi rh}{k} \right){{\psi }^{\left( 1 \right)}}\left( \frac{r}{k} \right)}+\frac{1}{2{{\pi }^{2}}{{k}^{2}}}\sum\limits_{r=1}^{k-1}{\cos \left( \frac{2\pi rh}{k} \right)\zeta '\left( 2,\frac{r}{k} \right)}\\&+\frac{1}{{{k}^{2}}}\zeta '\left( -1 \right) \\ \end{align}$$ From this, finally: $$\begin{align} S\left( m \right)&=\frac{1}{2}+\log \left( \frac{\sqrt{2m}\Gamma \left( 1+\frac{1}{2m} \right)\Gamma \left( 1-\frac{1}{2m} \right)}{\sqrt{\pi \Gamma \left( 1+\frac{1}{m} \right)\Gamma \left( 1-\frac{1}{m} \right)}} \right)\\&+\frac{1}{4\pi m}\sum\limits_{r=1}^{m-1}{\sin \left( \frac{2\pi r}{m} \right){{\psi }^{\left( 1 \right)}}\left( \frac{r}{m} \right)}-\frac{1}{4\pi m}\sum\limits_{r=1}^{2m-1}{\sin \left( \frac{\pi r}{m} \right){{\psi }^{\left( 1 \right)}}\left( \frac{r}{2m} \right)}\end{align}$$

Simple example: m=2 $$S\left( 2 \right)=\frac{1}{2}+\log \left( \frac{2\sqrt{2}\Gamma \left( \frac{5}{4} \right)\Gamma \left( \frac{3}{4} \right)}{\pi } \right)-\frac{1}{8\pi }\left\{ {{\psi }^{\left( 1 \right)}}\left( \frac{1}{4} \right)-{{\psi }^{\left( 1 \right)}}\left( \frac{3}{4} \right) \right\}$$ Simplifying $$S\left( 2 \right)=\frac{1}{2}-\frac{1}{8\pi }\left\{ {{\psi }^{\left( 1 \right)}}\left( \frac{1}{4} \right)-{{\psi }^{\left( 1 \right)}}\left( \frac{3}{4} \right) \right\}=\frac{1}{2}-\frac{2G}{\pi }$$ where G is Catalan’s constant.

References: Srivastava, H. M., Choi, J. Zeta and q-Zeta Functions and Associated Series and Integrals (Elsevier 2012)

I didn't want to rain on mathstackuser12's parade, but now that the bounty has been awarded, here's a simpler approach using the theory of the Barnes G function. (Mathstackuser12 seems to have reinvented a lot of it.) As that answer notes, use the log representation of arccoth. Sum in pairs up to $2N,$ where $N$ will be driven to $\infty.$ The sum over $(-1)^n \cdot (-1)$ will then be zero (it is included so the summand makes sense.) Then

$$S:=\sum_{n=1}^\infty (-1)^n\big(m \, n \, \text{arccoth}(m \, n) -1 \big) = \lim_{N \to \infty} \frac{m}{2}\log\Big( \prod_{n=1}^{2N} \Big(\frac{1+1/(m \, n)}{1-1/(m\,n)} \Big)^{(-1)^n n } \Big)$$ This is done because Adamchik has proven, (Proposition 5 of reference below)

$$ \lim_{N \to \infty} \prod_{n=1}^{2N} \Big(1+2x/n \Big)^{(-1)^{n+1} n }=\frac{e^{-x}\Gamma(x+1/2)}{\Gamma(1/2)} \Big(\frac{G(x+1/2)}{G(x+1)G(1/2)}\Big)^2$$

where $G$ is the Barnes G function.

Apply this formula to numerator and denominator of right hand side of top formula with $x=1/(2m)$ and $x=-1/(2m):$

$$S=\frac{m}{2} \log\Bigg( \Big( \frac{ G(1/2-1/(2m))G(1+1/(2m))}{G(1/2+1/(2m))G(1-1/(2m))} \Big)^2 \frac{\Gamma(1/2-1/(2m))}{\Gamma(1/2+1/(2m))} \exp{(1/m)} \Bigg)$$

Now use 2 functional equations of the Barnes G

$$\frac{G(1-x)}{G(1+x)} =\left(\frac{\sin{\pi x}}{\pi}\right)^x \exp{\left(\frac{\text{Cl}_2(2 \pi x)}{2\pi}\right) }\, , \quad G(1+x)=G(x)\Gamma(x)$$ where Cl$_2(x)$ is the Clausen function mentioned in the statement of the problem, and $0<x<1.$ The second equation is needed to to shift the ratio G(1/2-x)/G(1/2+x), which is the first factor in right hand side of the penultimate equation, so that it looks like the first functional equation. Finally, some algebra and a gamma function identity puts the equation in the form

$$S=\frac{1}{2} + \frac{1}{2}\log\big({\cot(\frac{\pi}{2m})}\big) + \frac{m}{2\pi}\big(\text{Cl}_2(\pi/m- \pi)-\text{Cl}_2(\pi/m) \big) $$

Besides an economy of getting to the answer, the answer is stated in terms of Clausen functions, which is what the OP found in his examples. Note also that there is no need for $m$ being an integer. (Numerical checks have verified it.)

Ref.: 'Multiple Gamma Function and Its Application to Computation of Series,' V.S. Adamchik, arXiv:math.CA/0308074v1 7 Aug 2003.

Assume that $m >1$.

Similar to my answer for the non-alternating version of the series, we can exploit the fact that $\csc(\pi z)$ has the Laurent series expansion $$\csc(\pi z) = \frac{1}{\pi z} + \frac{2}{\pi z} \sum_{n=1}^{\infty} \eta(2n)z^{2n}, \quad 0 <|z| < 1, $$ where $\eta(z)$ is the Dirichlet eta function.

Then again using the the Laurent series expansion of $\operatorname{arcoth}(z)$ at $\infty$, we get $$ \begin{align} S&= \sum_{n=1}^{\infty} (-1)^{n} \left(mn \operatorname{arcoth}(mn)-1 \right) \\ &= \sum_{n=1}^{\infty} (-1)^{n} \left(mn \sum_{k=0} \frac{1}{(2k+1)(mn)^{2k+1}} -1\right) \\ &= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{ (-1)^{n}}{(2k+1)(mn)^{2k}} \\ &= \sum_{k=1}^{\infty} \frac{1}{(2k+1)m^{2k}}\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2k}} \\ &= -\sum_{k=1}^{\infty} \frac{\eta(2k)}{(2k+1)m^{2k}} \\ &= \frac{m}{2} \int_{0}^{1/m} \mathrm dx - \frac{m \pi}{2} \int_{0}^{1/m} x\csc(\pi x) \, \mathrm dx \\ &= \frac{1}{2} + \frac{mx}{2} \log \left(\cot \frac{\pi x}{2} \right)\Bigg|_{0}^{1/m} - \frac{m}{2} \int_{0}^{1/m} \log \left(\cot \frac{\pi x}{2} \right) \, \mathrm dx \\ &= \frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right)- \frac{m}{2} \int_{0}^{1/m} \log \left(\cos \frac{\pi x}{2} \right) \, \mathrm dx + \frac{m}{2} \int_{0}^{1/m} \log \left(\sin \frac{\pi x}{2} \right) \, \mathrm dx \\ &= \frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right) - \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(\cos \frac{u}{2} \right) \, \mathrm du + \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(\sin \frac{u}{2} \right) \, \mathrm du \\ &= \frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right)- \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(2 \cos \frac{u}{2} \right) \, \mathrm du + \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(2 \sin \frac{u}{2} \right) \, \mathrm du \\ &=\frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right) - \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(2 \cos \frac{u}{2} \right) \, \mathrm du - \frac{m}{2 \pi} \operatorname{Cl}_{2} \left(\frac{\pi}{m} \right). \end{align}$$

The substitution $w= \pi -u$ shows that $$-\int_{0}^{\pi/m} \log \left(2 \cos \frac{u}{2} \right) \, \mathrm du = - \operatorname{Cl}_{2} \left(\pi - \frac{\pi}{m} \right)=\operatorname{Cl}_{2} \left(\frac{\pi}{m}- \pi \right). $$

Therefore, $$\sum_{n=1}^{\infty} (-1)^{n} \left(mn \operatorname{arcoth}(mn)-1 \right) = \frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right) + \frac{m}{2 \pi} \left(\operatorname{Cl}_{2} \left(\frac{\pi}{m}- \pi \right) - \operatorname{Cl}_{2} \left(\frac{\pi}{m} \right) \right), $$ which is the same result that skbmoore got.