<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] 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 $$ \begin{align} S&=  \sum_{n=1}^{\infty} (-1)^{n} \left(mn \operatorname{arcoth}(mn)-1 \right) \\ &= \sum_{n=1}^{\infty} (-1)^{n} \left(mn \sum_{k=0}^{\infty} \frac{1}{(2k+1)(mn)^{2k+1}} -1\right) \\ &= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{ (-1)^{n}}{(2k+1)(mn)^{2k}} \\ &=  \sum_{k=1}^{\infty} \frac{1}{(2k+1)m^{2k}}\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2k}} \\ &=  -\sum_{k=1}^{\infty} \frac{\eta(2k)}{(2k+1)m^{2k}} \\ &= \frac{m}{2} \int_{0}^{1/m} \mathrm dx - \frac{m \pi}{2} \int_{0}^{1/m} x\csc(\pi x) \, \mathrm dx \\ &= \frac{1}{2} + \frac{mx}{2} \log \left(\cot \frac{\pi x}{2} \right)\Bigg|_{0}^{1/m} - \frac{m}{2} \int_{0}^{1/m} \log \left(\cot \frac{\pi x}{2} \right) \, \mathrm dx \\ &= \frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right)- \frac{m}{2} \int_{0}^{1/m} \log \left(\cos \frac{\pi x}{2} \right) \, \mathrm dx + \frac{m}{2} \int_{0}^{1/m} \log \left(\sin \frac{\pi x}{2} \right) \, \mathrm dx \\ &= \frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right) - \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(\cos \frac{u}{2} \right) \, \mathrm du + \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(\sin \frac{u}{2} \right) \, \mathrm du \\ &= \frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right)- \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(2 \cos \frac{u}{2} \right) \, \mathrm du + \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(2 \sin \frac{u}{2} \right) \, \mathrm du  \\ &=\frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right) - \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(2 \cos \frac{u}{2} \right) \, \mathrm du  - \frac{m}{2 \pi} \operatorname{Cl}_{2} \left(\frac{\pi}{m} \right). \end{align}$$
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.
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