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I'm looking for a closed-form for the following sum:

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

In a previous question of mine, the following similar sum was determined:

$$\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)=\frac{1}{2}+\frac{1}{2} \log \left|\cot\left(\frac{\pi}{2m}\right)\right|+\frac{m}{2\pi} \left(\frac{1}{2}\text{Cl}_2\left(\frac{2\pi}{m}\right)-2\text{Cl}_2\left(\frac{\pi}{m}\right)\right)$$ for $|m|>1,$ where $\text{Cl}_2$ is the Clausen function of order 2.

I determined the following closed-forms for example as mentioned in that question:

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

I suspect a similar method to that used by @skbmoore involving the Barnes G function might be applicable. Any help would be much appreciated.


Here is my attempt:

$$S(m) := \sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right) = \frac{m}{2} \log \left( \prod_{n=1}^{\infty} \frac{1}{e} \left(\frac{1+1/(m n)}{1-1/(m n)}\right)^{n}\right)\\ = \frac{1}{2} + \frac{m}{2} \zeta' \left(-1,1-\frac{1}{m}\right)-\frac{m}{2}\zeta' \left(-1,1+\frac{1}{m}\right)+\frac{1}{2} \zeta' \left(0,1-\frac{1}{m}\right)+\frac{1}{2} \zeta' \left(0,1+\frac{1}{m}\right) \\ =\frac{1}{2} + \frac{m}{2} \left(\zeta' \left(-1,1-\frac{1}{m}\right)-\zeta' \left(-1,1+\frac{1}{m}\right)\right) \\+ \frac{1}{2} \ln\left( \Gamma \left(1-\frac{1}{m}\right)\Gamma \left(1+\frac{1}{m}\right)\right) - \frac{1}{2}\ln (2\pi) \\=\frac{1}{2} + \frac{m}{2} \left(\zeta' \left(-1,1-\frac{1}{m}\right)-\zeta' \left(-1,1+\frac{1}{m}\right)\right) + \frac{1}{2} \ln\left( \frac{\pi}{m} \csc \left(\frac{\pi}{m}\right)\right) - \frac{1}{2}\ln (2\pi)$$

However, I would like to write the solution in terms of the Clausen function if possible, like in the alternating sum, but I cannot see how to do that.

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    $\begingroup$ You said a similar method may be used, but did you try it? $\endgroup$ Dec 28, 2021 at 0:08

2 Answers 2

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EDIT: There was a question asked a long time ago about the related infinite series $$\sum_{n=1}^{\infty} \left( n \operatorname{arccot}(n)-1 \right). $$

The approach used in the accepted answer is similar to the approach I used for this question.


For $|x| >1 $, the inverse hyperbolic cotangent function has the Laurent series representation $$\operatorname{arcoth}(x) = \frac{1}{2} \log \left(\frac{1+ \frac{1}{x}}{1- \frac{1}{x}} \right) = \frac{1}{2} \log \left[ \left(1+ \frac{1}{x} \right) - \log\left(1- \frac{1}{x} \right) \right]= \sum_{n=0}^{\infty}\frac{1}{(2n+1)x^{2n+1}}. $$ And for $0 < |x| < 1$, $\cot (\pi x)$ has the Laurent series representation $$\cot(\pi x) = \frac{1}{\pi x} - \frac{2}{\pi x} \sum_{k=1}^{\infty}\zeta(2k) x^{2k}. $$

Therefore, for $m>1$, we have $$ \begin{align} \sum_{n=1}^{\infty} \left(mn \operatorname{arcoth} (mn)-1 \right) &= \sum_{n=1}^{\infty} \left(mn \sum_{k=0}^{\infty} \frac{1}{(2k+1) (mn)^{2k+1}} -1\right) \\ &= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{(2k+1)(mn)^{2k}} \\ &= \sum_{k=1}^{\infty}\frac{1}{(2k+1)m^{2k}} \sum_{n=1}^{\infty} \frac{1}{n^{2k}} \\ &= \sum_{k=1}^{\infty}\frac{\zeta(2k)}{(2k+1)m^{2k}} \\ &= \frac{m}{2} \int_{0}^{1/m} \, \mathrm dx - \frac{m \pi}{2} \int_{0}^{1/m} x \cot (\pi x) \, \mathrm dx \\ &= \frac{1}{2} - \frac{m x \log(\sin \pi x)}{2}\Bigg|_{0}^{1/m} + \frac{m}{2} \int_{0}^{1/m} \log(\sin \pi x) \, \mathrm dx \\ &= \frac{1}{2} - \frac{\log \left(\sin \frac{\pi}{m} \right)}{2} + \frac{m}{4 \pi} \int_{0}^{2 \pi /m} \log \left(\sin \frac{ u}{2}\right) \, \mathrm du \\ &= \frac{1}{2} - \frac{\log \left(\sin \frac{\pi}{m} \right)}{2} + \frac{m}{4 \pi}\int_{0}^{2 \pi /m} \log \left(2 \sin \frac{u}{2}\right) \, \mathrm du - \frac{\log 2}{2} \\ &= \frac{1}{2} - \frac{\log \left(\sin \frac{\pi}{m} \right)}{2} - \frac{m \operatorname{Cl}_{2} \left(\frac{2 \pi}{m} \right)}{4 \pi } - \frac{\log 2}{2}. \end{align} $$

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    $\begingroup$ Wow, I really like how simple and quick this method gets to the answer! (+1) $\endgroup$
    – KStarGamer
    Jan 13, 2022 at 22:44
  • $\begingroup$ Remarkably straightforward - your answers are always very educational for me! $\endgroup$
    – FShrike
    Jan 13, 2022 at 23:07
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To get your result in terms of the Clausen function of order $2$, we can use the identity $$ \begin{align} \zeta'(-1,x) &= -\log G(x+1) + x \log \Gamma (x) + \zeta'(-1) \\ &= - \log \Gamma(x) - \log G(x) + x \log \Gamma(x) + \zeta'(-1) \\ &= - \log G(x) + (x-1) \log \Gamma(x) + \zeta'(-1), \end{align}$$

which holds for $x>0$,

and the Barnes G reflection formula $$\log \left( \frac{G(1-x)}{G(1+x)} \right)= x\log \left(\frac{\sin \pi x}{\pi} \right) + \frac{\operatorname{Cl}_{2}(2 \pi x)}{2 \pi}, \quad 0 < x< 1, $$ which was used in skbmoore's answer to your previous question.

Then $$ \begin{align} \small\zeta' \left(-1, 1- \frac{1}{m}\right)- \zeta' \left(-1, 1+ \frac{1}{m}\right) &= \small- \log \left(\frac{G \left(1-\frac{1}{m}\right)}{G \left(1+ \frac{1}{m}\right)} \right) - \frac{1}{m} \log \left(\Gamma \left(1+ \frac{1}{m} \right) \Gamma \left(1- \frac{1}{m} \right) \right) \\ &= \small - \log \left(\frac{G \left(1-\frac{1}{m}\right)}{G \left(1+ \frac{1}{m}\right)} \right) - \frac{1}{m} \log \left(\frac{1}{m}\Gamma \left(\frac{1}{m}\right) \Gamma \left(1- \frac{1}{m} \right) \right) \\ &= \small- \frac{1}{m}\log \left(\frac{\sin \left(\frac{\pi}{m} \right)}{\pi} \right) - \frac{\operatorname{Cl}_{2} \left(\frac{2 \pi}{m} \right)}{2 \pi}- \frac{\log\left(\frac{\pi}{m} \csc\left(\frac{\pi}{m}\right)\right)}{m}. \end{align}$$

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  • $\begingroup$ Ah the first identity was what I was missing. Thanks! $\endgroup$
    – KStarGamer
    Jan 10, 2022 at 7:27

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