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Do you know of a way to compute the following infinite series in a closed form expression

$$ \sum_{n=1}^{+\infty} \frac{1}{\sinh(n\,\alpha)}\,, $$

where $\alpha>0$.

Thanks in advance for the help!

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It's the $q$-digamma function $\psi_q(z)$: $$\psi_q(z) = \frac d {dz} \ln \Gamma_q(z) = -\ln(1 - q) - \ln(q) \sum_{k \geq 0} \frac 1 {1 - q^{-k - z}}, \\ \sum_{n \geq 1} \operatorname{csch} \alpha n = 2 \sum_{n \geq 1} \sum_{k \geq 0} e^{-\alpha n (2 k + 1)} = 2 \sum_{k \geq 0} \frac 1 {e^{\alpha (2 k + 1)} - 1} = \\ -\frac 1 \alpha \left( \psi_{\exp(-2 \alpha)} {\left( \frac 1 2 \right)} + \ln \left( 1 - e^{-2 \alpha} \right) \right).$$

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  • $\begingroup$ Fantastic! This is exactly what I wanted! Thanks! $\endgroup$ – user12588 May 31 at 20:52
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This is an example of a Lambert series and has no closed form that I know of but there is a power series expansion. Let

$$ f(\alpha) := \sum_{n=1}^{\infty} \frac{1}{\sinh(n\,\alpha)} = \sum_{n=1}^{\infty} \frac{2}{e^{n\alpha}-e^{-n\alpha}} \tag1 $$

where real part of $\,\alpha>0.\,$ If $\,\alpha = -\log q,\,$ then consider the power series

$$ \frac12 f(\alpha) = \sum_{n=1}^{\infty} a_n q^n = q + q^2 + 2q^3 + q^4 + 2q^5 + 2q^6 + \cdots\tag2 $$

where $a_n$ is OEIS sequence A001227 "Number of odd divisors of n". You can read the sequence entry and it may give further information you can use. Note that the radius of convergence for $q$ is $1$ and the unit circle is a natural boundary. This corresponds to real part of $\alpha>0$.

EDIT As the answer by 'Maxim' notes, there exists a closed form using the $q$-digamma function given by $\,f(\alpha) = (\psi_{q^2}(1/2)+\log(1-q^2))/\log(q).$ Read the q-Polygamma page for details.

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