# 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.

• Just a thought: The sum can be also written as $$\sum_{n=1}^{\infty} \chi(n)\log(1 - e^{-2\pi n}),$$ where $\chi : \Bbb{Z} \to \Bbb{C}$ is the Dirichlet character to the modulus $5$ specified by $\chi(0) = 0$, $\chi(1) = \chi(4) = 1$ and $\chi(2) = \chi(3) = -1$. It may be a mere coincidence, but is it possible that this relation can be utilized somehow? Commented Dec 30, 2016 at 19:28
• @SangchulLee: that is an interesting observation, probably the Chowla-Selberg formula (en.wikipedia.org/wiki/Chowla%E2%80%93Selberg_formula) is useful here. Commented Dec 30, 2016 at 19:53
• You always post interesting problems. Commented Dec 30, 2016 at 20:18

$$\begin{eqnarray*}\sum_{n\geq 1}\frac{e^{2n\pi}}{n(e^{10n\pi}-1)}&=&\sum_{n\geq 1}\left(\frac{e^{-8\pi n}}{n}+\frac{e^{-18\pi n}}{n}+\frac{e^{-28\pi n}}{n}+\ldots\right)\\&=&-\log(1-e^{-8\pi})-\log(1-e^{-18\pi})-\ldots\\&=&\log\prod_{k\geq 0}\frac{1}{1-(e^{-\pi})^{10k+8}}\end{eqnarray*}$$ 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.
• See this post for value of $R(e^{-2\pi})$: paramanands.blogspot.com/2013/09/… Commented Jan 1, 2017 at 13:31