$$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n}\left(\frac{e^{-\pi n}}{1+e^{-\pi n}}\right)=\frac{\pi-5\ln2}{8} \tag{1}$$
I am searching for an elementary way to prove this considering the simplicity of the result.
This is how I basically came up with it.
Consider the Function: $$\delta(x)=\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{x^n}{1-x^n}\right)$$
Now it can be shown that: $$-2\delta\left(x^{4}\right)+3\delta\left(x^{2}\right)-\delta\left(x\right)=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n}\left(\frac{x^{n}}{1+x^{n}}\right) \tag{2}$$
What's interesting here is that we have Closed Form Values of the Function Evaluated at $\left(e^{-\pi}\right)^1$, $\left(e^{-\pi}\right)^2$, $\left(e^{-\pi}\right)^4$ which were given by Sir Ramanujan as far as I know.
These are,
$$\delta(e^{-\pi})=\frac{7}{8}\ln2+\frac{3}{4}\ln\pi-\frac{1}{24}\pi-\ln\Gamma\left(\frac{1}{4}\right)$$ $$\delta(e^{-2\pi})=\ln2+\frac{3}{4}\ln\pi-\frac{1}{12}\pi-\ln\Gamma\left(\frac{1}{4}\right)$$ $$\delta(e^{-4\pi})=\frac{11}{8}\ln2+\frac{3}{4}\ln\pi-\frac{1}{6}\pi-\ln\Gamma\left(\frac{1}{4}\right)$$
Now putting $e^{-\pi}$ in $(2)$ Results in $(1)$.
Is there an elementary way to prove this considering the Simplicity of the Result.
