Given $\varphi$ is golden ratio, how do I prove this $\sum \limits_{j=1}^{\infty}\frac{(1-\varphi)^j}{j^2}\cos{\frac{3j\pi}{5}}=\frac{\pi^2}{100}$?

Question

Given $ \varphi$ is golden ratio, how do I prove this: $ \displaystyle \tag*{}\sum \limits_{j=1}^{\infty}\frac{(1-\varphi)^j}{j^2}\cos{\frac{3j\pi}{5}}=\frac{\pi^2}{100}$

My approach:

We can reduce the sum into simpler parts by using the property of golden ratio, we have:

$\displaystyle \tag*{} \phi^2 - \phi - 1 = 0 \Leftrightarrow 1- \phi = - \dfrac{1}{\phi}$

And I also found the values of $\cos \dfrac{3j\pi}{5}$ in terms of $\varphi$:

$\displaystyle \tag*{} \begin{align} \cos \dfrac{3\pi}{5} &= \dfrac{-1}{2 \varphi} \\\\ \cos \dfrac{6\pi}{5} &= \dfrac{-\varphi}{2} \\\\ \cos \dfrac{9\pi}{5} &= \dfrac{\varphi}{2} \\\\ \cos \dfrac{12\pi}{5} &= \dfrac{1}{2 \varphi} \\\\ \cos \dfrac{15\pi}{5} &= {-1} \end{align}$

And this repeats, periodically with alternate opposite signs. I don't know how to connect these information I found to prove the question. Maybe my approach is wrong. Any help would be appreciated. Thanks.

Answer 1

The point is the functional equation $$Li_2(1-z)+Li_2(1-1/z)=-\frac12 \log^2 z$$ of the dilogarithm $$Li_2(z)=\sum_{k\ge 1} \frac{z^k}{k^2}$$ which follows from $z \, Li_2'(z)=-\log(1-z)$,

and that $$1-z= (1-\varphi)e^{3i\pi/5} \implies z=e^{i\pi/5}$$

So $$\sum_{k\ge 1} \frac{(1-\varphi)^k}{k^2}\cos{\frac{3k\pi}{5}}=\frac{-1}4 \log^2 e^{i\pi/5} = \frac{\pi^2}{100}$$

Answer 2

Hint

For $x \in \mathbb R$ $$\cos x =\frac{e^{ix}+e^{-ix}}{2}$$

Now let $$f(x)= \sum_{n=1}^\infty \frac{x^n}{n^2}.$$

We have $f(0)=0$ and $$f^\prime(x)=\sum_{n=1}^\infty \frac{x^{n-1}}{n}=-\frac{1}{x}\ln(1-x)$$