<s>[INST] General form for $\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$ I'm wondering if there is a general form for the following sum:

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

I have obtained the following closed-forms for these special cases:

Where $G$ is Catalan's constant and $\text{Cl}_2$ is the Clausen function of order 2.
$$\sum_{n=1}^{\infty}(-1)^n \left(2n \, \text{arccoth} \, (2n) - 1\right)  = \frac{1}{2} - \frac{2G}{\pi}$$
$$\sum_{n=1}^{\infty} (-1)^{n}\left( 3n \, \text{arccoth} \, (3n)-1\right)  = \frac{1}{2} - \frac{5}{2\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{3}\right) + \frac{1}{4} \ln (3)$$
$$\sum_{n=1}^{\infty} (-1)^n \left(4n \, \text{arccoth} \, (4n) - 1\right) = \frac{1}{2}+ \frac{G}{\pi} - \frac{4}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{1}{4} \ln \left(3- 2\sqrt{2}\right)$$
etc.

Given that $G = \text{Cl}_2 \left(\frac{\pi}{2}\right)$, I am curious to know if the general sum is expressible in terms of the Clausen function. These above sums were determined by using the Mittag-Leffler expansion of $\csc (z)$, i.e $\csc(z) = \frac{1}{z} + 2z \sum_{n=1}^{\infty} (-1)^n \frac{1}{z^2 - \left(\pi n\right)^2}$ and substituting it into the integral $\int_{0}^{\pi/m} x \csc (x) \, dx$
If one uses the following other method, we can determine the odd and even terms of the sums

$$\sum_{n=1}^{\infty} \left( 4n \, \text{arccoth} \, (4n)-1\right) = \frac{1}{2} - \frac{G}{\pi}- \frac{1}{4} \ln (2)$$
$$\sum_{n=1}^{\infty} \left( (4n-2) \, \text{arccoth} \, (4n-2) - 1\right) = \frac{G}{\pi} - \frac{1}{4} \ln (2)$$
$$\sum_{n=1}^{\infty} \left(6n \, \text{arccoth} \, (6n) - 1\right) = \frac{1}{2} - \frac{3}{2\pi} \, \text{Cl}_2 \left( \frac{\pi}{3}\right)$$
$$\sum_{n=1}^{\infty} \left( (6n-3) \, \text{arccoth} \, (6n-3) - 1\right) = \frac{1}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{3}\right) - \frac{1}{4} \ln(3)$$
$$\sum_{n=1}^{\infty} \left( 8n \, \text{arccoth} \, (8n) - 1\right) = \frac{1}{2} - \frac{2}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{1}{4} \ln (2-\sqrt{2})$$
$$\sum_{n=1}^{\infty} \left( (8n-4) \, \text{arccoth} \, (8n-4) - 1\right) = \frac{2}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{G}{\pi} - \frac{1}{4} \ln(2+\sqrt{2})$$

EDIT

I am now interested in
$$\sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right)$$
for $|m|>1$ as asked here.


Here is how I originally proved it for the odd terms:
For the first odd term sum, begin with the known result $$2G = \int_{0}^{\frac{\pi}{2}} x \csc (x) \, dx$$
then integrate by parts to get:
$$2G = \int_{0}^{\frac{\pi}{2}} \ln \left(\cot (x) + \csc (x) \right) \, dx = \int_{0}^{\frac{\pi}{2}} \ln (\cos (x) + 1) \, dx - \int_{0}^{\frac{\pi}{2}} \ln (\sin (x)) \,dx$$
Then use the well-known result $\int_{0}^{\frac{\pi}{2}} \ln (\sin(x)) \,dx = - \frac{\pi}{2} \ln (2)$ and use the identity $\cos (x) + 1 = 2 \cos^2\left( \frac{x}{2}\right)$ and make the substitution $\frac{x}{2} = u$.
$$\implies 2G - \pi \ln (2) = 4 \int_{0}^{\frac{\pi}{4}} \ln (\cos (u)) \, du$$
Now use the Weierstrass product for $\cos (z)$, namely $\cos(z) = \prod_{n=1}^{\infty} \left(1-\frac{4z^2}{\pi^2 (2n-1)^2}\right)$ to obtain:
$$2G - \pi \ln (2) = 4 \sum_{n=1}^{\infty} \int_{0}^{\frac{\pi}{4}} \ln \left( 1-\frac{4u^2}{\pi^2 (2n-1)^2}\right) \, du$$
After integrating, obtain $\pi \sum_{n=1}^{\infty} \ln \left(1-\frac{1}{4(1-2n)^2}\right) = -\frac{\pi}{2} \ln (2)$ and the result quickly follows. The other odd sums are the same idea. The even sums just comes from combining the two results from the alternating sum and the odd term sum.
 [/INST] 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.
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