<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] 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 $$f^\prime(x)=\sum_{n=1}^\infty \frac{x^{n-1}}{n}=-\frac{1}{x}\ln(1-x)$$
</s>