Prove $\ln(2^5)-\pi=8\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{e^{n\pi}+1}-\frac{1}{e^{2n\pi}+1}\right)$ We took this idea from Simon Plouffe see here
$$\ln(2^5)-\pi=8\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{e^{n\pi}+1}-\frac{1}{e^{2n\pi}+1}\right)$$

Can anyone prove this identiy?
We found this identity via a sum calculator by varying Simon Plouffe identities

We can calculate the sum $$\sum_{n = 1}^{\infty}\frac{1}{n(e^{nx} + 1)}$$ as follows.
Let $q = e^{-x}$ and then we have
\begin{align}
F(q) &= \sum_{n = 1}^{\infty}\frac{1}{n(e^{nx} + 1)}\notag\\
&= \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 + q^{n})}\notag\\
&= \sum_{n = 1}^{\infty}\frac{1}{n}\sum_{k = 1}^{\infty}(-1)^{k - 1}q^{kn}\notag\\
&= \sum_{k = 1}^{\infty}(-1)^{k - 1}\sum_{n = 1}^{\infty}\frac{q^{kn}}{n}\notag\\
&= \sum_{k = 1}^{\infty}(-1)^{k}\log(1 - q^{k})\notag\\
&= \sum_{k \text{ even}}\log(1 - q^{k}) - \sum_{k \text{ odd}}\log(1 - q^{k})\notag\\
&= 2\sum_{k \text{ even}}\log(1 - q^{k}) - \sum_{k = 1}^{\infty}\log(1 - q^{k})\notag\\
&= 2\sum_{k = 1}^{\infty}\log(1 - q^{2k}) - \sum_{k = 1}^{\infty}\log(1 - q^{k})\notag\\
&= 2f(q^{2}) - f(q)\notag
\end{align}
where we have from the linked answer (the $f(q)$ of this answer is same as $-a(q)$ of the linked answer, also note that the linked answer proves another formula of Plouffe for $\pi$) $$f(q) = \sum_{k = 1}^{\infty}\log(1 - q^{k}) = \frac{\log k}{12} + \frac{\log k'}{3} + \frac{\log 2}{3} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{24K}\tag{1}$$ and $$f(q^{2}) = \frac{\log(kk')}{6} + \frac{\log 2}{6} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{12K}\tag{2}$$ and $$f(q^{4}) = \frac{1}{12}\log(k^{4}k') - \frac{\log 2}{6} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{6K}\tag{3}$$ From the above equations we get $$F(q) = \frac{\log k}{4} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{8K}\tag{4}$$ and $$F(q^{2}) = \frac{\log k}{2} - \frac{\log 2}{2} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{4K}\tag{5}$$ The sum in your question is $$8(F(q) - F(q^{2}))$$ where $q = e^{-\pi}$ and this is clearly equal to $$8\left(-\frac{\log k}{4} + \frac{\log 2}{2} - \frac{\pi K'}{8K}\right)$$ and we note that $k = 1/\sqrt{2}$ and $K' = K$ for $q = e^{-\pi}$. Hence the desired sum is $$8((5/8)\log 2 - (\pi/8)) = 5\log 2 - \pi$$

Most of the sums which involve the expressions of type $e^{n\pi}$ I try to express them as function of $q = e^{-\pi}$ and hope that the result series / product in $q$ can be related to the elliptic integrals $K$ and modulus $k$. If this is possible (like here) then the problem of dealing with infinite series/product is transformed into dealing with finite expressions consisting of $k, K, \pi$ and simple algebraic manipulations on them. This may not work always (or may not be convenient every time), but appears simpler to me when it works compared to general techniques for summing infinite series.
