<s>[INST] Given $\varphi$ is golden ratio, how do I prove this $\sum \limits_{j=1}^{\infty}\frac{(1-\varphi)^j}{j^2}\cos{\frac{3j\pi}{5}}=\frac{\pi^2}{100}$? 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.
 [/INST] 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 $$\sum_{k\ge 1} \frac{(1-\varphi)^k}{k^2}\cos{\frac{3k\pi}{5}}=\frac{-1}4 \log^2 e^{i\pi/5} = \frac{\pi^2}{100}$$
</s>