Prove that: $\sum\limits_{n=1}^{\infty}{e^{2n\pi}-e^{4n\pi}-e^{6n\pi}+e^{8n\pi}\over n(1-e^{10n\pi})}=e^{2\pi\over 5}\ln{(\sqrt{\phi+2}-\phi)}$ Show that
$$\sum_{n=1}^{\infty}{e^{2n\pi}-e^{4n\pi}-e^{6n\pi}+e^{8n\pi}\over n(1-e^{10n\pi})}=e^{2\pi\over 5}\ln{(\sqrt{\phi+2}-\phi)}$$
$\phi$; golden ratio
My try:
$$\sum_{n=1}^{\infty}{e^{2n\pi}\over n(1-e^{10n\pi})}-\sum_{n=1}^{\infty}{e^{4n\pi}\over n(1-e^{10n\pi})}-\sum_{n=1}^{\infty}{e^{6n\pi}\over n(1-e^{10n\pi})}+\sum_{n=1}^{\infty}{e^{8n\pi}\over n(1-e^{10n\pi})}
$$
We can use, because the denumerator are same, but still numerator is a problem
$$-\ln{\left(\prod_{r=1}^{\infty}{n^r\over n^r-1}\right)}=\sum_{k=1}^{\infty}{1\over k}\left({1\over 1- n^k}\right)$$
$$-\ln{\left(\prod_{r=1}^{\infty}{e^{10r\pi}\over e^{10r\pi}-1}\right)}=\sum_{k=1}^{\infty}{1\over k}\left({1\over 1- e^{10k\pi}}\right)$$
The $e^{2n\pi}$ is missing From the numerator. I can't think of any other way of changing the formula to suit the problem above. Any help please.

$$\begin{eqnarray*}\sum_{n\geq 1}\frac{e^{2n\pi}}{n(e^{10n\pi}-1)}&=&\sum_{n\geq 1}\left(\frac{e^{-8\pi n}}{n}+\frac{e^{-18\pi n}}{n}+\frac{e^{-28\pi n}}{n}+\ldots\right)\\&=&-\log(1-e^{-8\pi})-\log(1-e^{-18\pi})-\ldots\\&=&\log\prod_{k\geq 0}\frac{1}{1-(e^{-\pi})^{10k+8}}\end{eqnarray*} $$
so the whole series equals, by setting $q=e^{-2\pi}$,
$$ \log\prod_{k\geq 0}\frac{(1-q^{5k+4})(1-q^{5k+1})}{(1-q^{5k+3})(1-q^{5k+2})} $$
that is related with the Rogers-Ramanujan continued fraction. 
The value of $R(e^{-2\pi})$ can be computed from the theory of modular forms and ensures the validity of the claim.
