Proving $\sum_{n=0}^{\infty}\frac{(\phi-1)^n}{(2n+1)^2}=\frac{\pi^2}{12}-\frac{3\ln^2(\phi)}{4}$ 
How do we prove this?
$$\sum_{n=0}^{\infty}\frac{(\phi-1)^n}{(2n+1)^2}=\frac{\pi^2}{12}-\frac{3\ln^2(\phi)}{4}$$
where $\phi:=\frac12(1+\sqrt{5})$ is the Golden Ratio.

My attempt:
\begin{align*}
\displaystyle\sum_{n=0}^{\infty}\frac{(\phi-1)^n}{(2n+1)^2}&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{\phi-1})^{2n}}{2n+1}\left(1-\frac{2n}{2n+1}\right)\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{\phi-1})^{2n}}{2n+1}-\displaystyle\sum_{n=0}^{\infty}\frac{(2n)(\sqrt{\phi-1})^{2n}}{(2n+1)^2}\\
&=J-I\\
\text{where}\\
J&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{\phi-1})^{2n}}{2n+1}\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n}}{2n+1}|_{x=\sqrt{\phi-1}}\\
\text{and}\\
\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n}}{2n+1}&=\frac{1}{x}\displaystyle\sum_{n=0}^{\infty}\displaystyle\int x^{2n}dx\\
&=\frac{1}{x}\displaystyle\int\frac{1}{1-x^2}dx\\
&=\frac{1}{2x}\ln\left(\frac{1+x}{1-x}\right)\\
\text{So}\\
J&=\frac{1}{2x}\ln\left(\frac{1+x}{1-x}\right)|_{x=\sqrt{\phi-1}}\\
&=\frac{1}{2\sqrt{\phi-1}}\ln\left(\frac{1+\sqrt{\phi-1}}{1-\sqrt{\phi-1}}\right) 
\end{align*}
But how do we calculate $I$ to get the result.
 A: Your formula has a typo, it should be :
$$\sum_{n=0}^\infty\frac{(\phi-1)^{\color{red}{2n+1}}}{(2n+1)^2}=\frac{\pi^2}{12}-\frac{3\ln^2(\phi)}{4}$$
It´s an instance of Legendre's Chi-Function evaluated at $x=\phi-1$
Here is the proof:
$$
\begin{aligned}
\mathrm{Li}_{2}(x)&=\sum_{n=1}^\infty\frac{x^n}{n^2}\\
&=\frac14\sum_{n=1}^\infty\frac{x^{2n}}{n^2}+\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)^2}\\
&=\frac14\mathrm{Li}_{2}(x^2)+\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)^2}\\
 \end{aligned}
$$
Hence
$$
\begin{aligned}
\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)^2}&=\mathrm{Li}_{2}(x)-\frac14\mathrm{Li}_{2}(x^2)\\
&=\frac12\left(\mathrm{Li}_{2}(x)-\mathrm{Li}_{2}(-x)\right)
 \end{aligned}
$$
Letting $x=\phi-1$ and noting that $\phi-1=\frac{1}{\phi}$ we obtain
$$
\begin{aligned}
\sum_{n=0}^\infty\frac{(\phi-1)^{2n+1}}{(2n+1)^2}&=\frac12\left(\mathrm{Li}_{2}\left(\frac{1}{\phi}\right)-\mathrm{Li}_{2}\left(-\frac{1}{\phi}\right)\right)\\
&=\frac12\left(\frac{\pi^2}{10}-\ln^2(\phi)+\frac{\pi^2}{15}-\frac{\ln^2(\phi)}{2}\right)\\
&=\frac{\pi^2}{12}-\frac{3\ln^2(\phi)}{4} \qquad \blacksquare
 \end{aligned}
$$
Where we used that
$
\mathrm{Li}_{2}(x)+\mathrm{Li}_{2}(-x)=\frac{1}{2} \mathrm{Li}_{2}\left(x^{2}\right)
$
and
$
\begin{aligned}
\mathrm{Li}_{2}\left(\frac{1}{\phi}\right) &=\frac{\pi^{2}}{10}-\ln ^{2} \phi \\
\mathrm{Li}_{2}\left(-\frac{1}{\phi}\right) &=\frac{\ln ^{2} \phi}{2}-\frac{\pi^{2}}{15} \\
\end{aligned}
$
A proof of these last two relations between dilogs and the golden ratio can be found here in my blog
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With the Golden Ratio $\ds{\phi \equiv {\root{5} + 1 \over 2}}$:
\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}{\pars{\phi - 1}^{n} \over
\pars{2n + 1}^{2}}}\qquad\qquad\qquad\qquad\qquad\qquad\pars{~\mbox{Note that}\ \phi - 1 = {1 \over \phi}~}
\\[5mm] & =
\root{\phi}\sum_{n = 0}^{\infty}{\pars{\phi^{-1/2}}^{2n + 1} \over
\pars{2n + 1}^{2}} = \root{\phi}
\sum_{n = 1}^{\infty}{\pars{\phi^{-1/2}}^{n } \over
n^{2}}\,{1 - \pars{-1}^{n} \over 2}
\\[5mm] = &\
{1 \over 2}\,\root{\phi}\bracks{\mrm{Li}_{2}\pars{1 \over \root{\phi}} -
\mrm{Li}_{2}\pars{-\,{1 \over \root{\phi}}}}
\approx 1.0919
\end{align}
