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.
Best Answer
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)