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?)

  • $\begingroup$ Do you mean $\tanh n$ or $\tanh (\pi n)$? $\endgroup$ – Olivier Oloa Jun 4 '16 at 8:40
  • 1
    $\begingroup$ 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 $\endgroup$ – Olivier Oloa Jun 4 '16 at 8:50
  • 4
    $\begingroup$ 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, $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n \tanh(\pi n)} = \frac{\pi}{6}+\frac14 \ln2.$$ $\endgroup$ – nospoon Jun 4 '16 at 9:12
  • 3
    $\begingroup$ An interesting paper (file.scirp.org/pdf/AM_2014102810361432.pdf) giving an answer to the OP question. Thanks to @nospoon. $\endgroup$ – Olivier Oloa Jun 4 '16 at 10:15
  • 2
    $\begingroup$ 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 $$\sum_{n,m=1}^{\infty} \frac{(-1)^{n+1}}{n^2+m^2}=\frac{\pi^2}{24}+\frac{\pi \ln2}{8}$$. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Tolaso Jun 4 '16 at 13:11
  • $\begingroup$ @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. $\endgroup$ – Paramanand Singh Jun 4 '16 at 13:14
  • $\begingroup$ 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 ... ! $\endgroup$ – Tolaso Jun 4 '16 at 13:19
  • $\begingroup$ @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. $\endgroup$ – Paramanand Singh Jun 4 '16 at 13:22
  • $\begingroup$ Good job there! (+1) $\endgroup$ – Olivier Oloa Jun 5 '16 at 8:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.