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?)
 A: 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.
