# Evaluation of $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \tanh n}$

Here is a series that arose while playing around with some differential equations.

$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \tanh n}$$

I have a feeling that it has a closed form. Although I am not able to attack it. For example an idea that could be promising would be to use the kernel $\pi \csc \pi z$ and integrate the function

$$f(z)=\frac{\pi \csc \pi z}{z \tanh z}$$

over a square, although I am unsure about the vertices. In the mean time Wolfram Alpha is unable to give a close form. Instead it returns $0.98903$ as an approximate result.

So, can anyone help me derive the closed form (if that eventually exists?)

• Do you mean $\tanh n$ or $\tanh (\pi n)$? – Olivier Oloa Jun 4 '16 at 8:40
• It seems that with $\tanh (\pi n)$ the given series is likely to belong to Plouffe like families ;) Now is there a closed form of it? I really don't know. vixra.org/abs/1409.0078 – Olivier Oloa Jun 4 '16 at 8:50
• We have $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \operatorname{coth}(n \pi x) = \frac{\pi x}{6}- \frac13 \ln \left( \frac{\theta_3(e^{\pi x}) \theta_4(e^{-\pi x})}{2 \theta_2^2(e^{-\pi x})}\right),$$ and in particular, $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n \tanh(\pi n)} = \frac{\pi}{6}+\frac14 \ln2.$$ – nospoon Jun 4 '16 at 9:12
• An interesting paper (file.scirp.org/pdf/AM_2014102810361432.pdf) giving an answer to the OP question. Thanks to @nospoon. – Olivier Oloa Jun 4 '16 at 10:15
• For the case with $\operatorname{coth}(\pi n)$ we can also write $$\frac{\pi}{n}\operatorname{coth}(\pi n)=\frac1{n^2}+2\sum_{m=1}^{\infty} \frac1{n^2+m^2},$$ i.e, it suffices to show that $$\sum_{n,m=1}^{\infty} \frac{(-1)^{n+1}}{n^2+m^2}=\frac{\pi^2}{24}+\frac{\pi \ln2}{8}$$. – nospoon Jun 4 '16 at 10:36

As noted in comments the sum $$\sum_{n = 1}^{\infty}\frac{(-1)^{n - 1}}{n\tanh n}$$ has no closed form except in the form of Jacobi's theta functions. I provide here an approach which evaluates the function $$F(x) = \sum_{n = 1}^{\infty}\frac{(-1)^{n - 1}}{n\tanh n x}$$ in terms of theta functions and then using this expression in theta functions we can get a value of $F(\pi)$ in closed form. Your questions asks for a closed form of $F(1)$ which does not seem possible.

Let $q = e^{-x}$ then we have \begin{align} F(x) &= \sum_{n = 1}^{\infty}\frac{(-1)^{n - 1}}{n}\cdot\frac{e^{nx} + e^{-nx}}{e^{nx} - e^{-nx}}\notag\\ &= \sum_{n = 1}^{\infty}\frac{(-1)^{n - 1}}{n}\cdot\frac{1 + q^{2n}}{1 - q^{2n}}\notag\\ &= \sum_{n = 1}^{\infty}\frac{(-1)^{n - 1}}{n}\cdot\left(1 + 2\frac{q^{2n}}{1 - q^{2n}}\right)\notag\\ &= \log 2 + 2\sum_{n = 1}^{\infty}\frac{(-1)^{n - 1}q^{2n}}{n(1 - q^{2n})}\notag\\ &= \log 2 + 2\sum_{n\text{ odd}}\frac{q^{2n}}{n(1 - q^{2n})} - 2\sum_{n\text{ even}}\frac{q^{2n}}{n(1 - q^{2n})}\notag\\ &= \log 2 + 2\sum_{n = 1}^{\infty}\frac{q^{2n}}{n(1 - q^{2n})} - 2\sum_{n = 1}^{\infty}\frac{q^{4n}}{n(1 - q^{4n})}\notag\\ &= \log 2 + 2a(q^{2}) - 2a(q^{4})\notag\\ \end{align} where $a(q^{2}), a(q^{4})$ are given in terms of elliptic integrals $K, K'$ and modulus $k$ by the equations: $$a(q^{2}) = -\frac{\log(kk')}{6} - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{12K}\tag{1}$$ and $$a(q^{4}) = -\frac{1}{12}\log(k^{4}k') + \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{6K}\tag{2}$$ and hence $$F(x) = \log 2 + 2\left(\frac{\log k}{6} - \frac{\log k'}{12} - \frac{\log 2}{3} + \frac{\pi K'}{12K}\right)$$ or $$F(x) = \frac{\log 2}{3} + \frac{\log k}{3} - \frac{\log k'}{6} + \frac{\pi K'}{6K}\tag{3}$$ and noting that $$k = \frac{\vartheta_{2}^{2}(e^{-x})}{\vartheta_{3}^{2}(e^{-x})}, k' = \frac{\vartheta_{4}^{2}(e^{-x})}{\vartheta_{3}^{2}(e^{-x})}$$ and $x = \pi K'/K$ we get $$F(x) = \frac{\log 2}{3} + \frac{x}{6} + \frac{1}{3}\log\frac{\vartheta_{2}^{2}(e^{-x})}{\vartheta_{3}(e^{-x})\vartheta_{4}(e^{-x})} = \frac{x}{6} - \frac{1}{3}\log\left(\frac{\vartheta_{3}(e^{-x})\vartheta_{4}(e^{-x})}{2\vartheta_{2}^{2}(e^{-x})}\right)\tag{4}$$ which is same as nospoon's formula in comments with $x$ in place of $\pi x$. If $x = \pi$ in my notation then $k = k' = 1/\sqrt{2}$ and then we obtain the sum $$\sum_{n = 1}^{\infty} \frac{(-1)^{n - 1}}{n \tanh n\pi} = \frac{\pi}{6} + \frac{\log 2}{4}\tag{5}$$

Note further that from the theory of modular equations it follows that if $K'/K = x/\pi = \sqrt{r}$ where $r$ is a positive rational number then the values of $k, k'$ are algebraic numbers and hence there is a closed form for $F(x) = F(\pi\sqrt{r})$ for all positive rational numbers $r$. Evaluating the closed form consisting of actual numbers becomes increasingly difficult (at least for hand calculation) as the numerator and denominator of $r$ increase in value.

• Marvellous! Brilliant! Thank you for your answer. I should have imagined that theta functions would come in play. Of course I realized that after nospoon's comment. Thanks again. – Tolaso Jun 4 '16 at 13:11
• @Tolaso: Whenver I see a series involving hyperbolic functions of argument $nx$ or $n\pi x$ for integer $n$ I sense that I can put $q = e^{-x}$ or $q = e^{-\pi x}$ and express the resulting $q$-series in terms of theta functions and finally reduce them to an expression $k, K, K'$. If the series is nice this works fine most of the times. – Paramanand Singh Jun 4 '16 at 13:14
• I guess have to learn a lot more for elliptic function, let alone theta Jacobi functions. It is a field that I have not investigated that much ... ! – Tolaso Jun 4 '16 at 13:19
• @Tolaso: You can search for elliptic integrals, theta functions, modular equations on my blog archive page paramanands.blogspot.com/p/archives.html and get all the introductory material on link between theta functions and elliptic integrals. – Paramanand Singh Jun 4 '16 at 13:22
• Good job there! (+1) – Olivier Oloa Jun 5 '16 at 8:11