Fibonacci sum for $\pi$: $\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}$

Question

Wikipedia's "List of formulae involving $\pi$" entry states that $$\sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}}=\frac{4\pi^2}{25\sqrt5}.$$ Why is this true?

If $\varphi=(1+\sqrt5)/2$ and $\psi=(1-\sqrt5)/2$, then we can rewrite this identity as $$F(\varphi)-F(\psi)=\frac{4\pi^2}{25}\quad\text{ where }F(x)=\sum_{n=1}^\infty\frac{x^{2n}}{n^2\binom{2n}{n}}.$$ $F(x)$ looks similar to the power series for arcsin.

Answer 1

$$F(x)=\sum_{n=1}^\infty\frac{x^{2n}}{n^2\binom{2n}{n}}=2 \left[\sin ^{-1}\left(\frac{x}{2}\right)\right]^2$$ makes $$F(\varphi)=\frac{9 \pi ^2}{50}\qquad \text{and}\qquad F(\psi)=\frac{\pi ^2}{50}$$ since $$\varphi=1-2 \cos \left(\frac{3 \pi }{5}\right)$$

Answer 2

In this answer, it is shown that $$ \frac{\arcsin(x)}{\sqrt{1-x^2}} =\sum_{k=0}^\infty\frac1{(2k+1)}\frac{4^k}{\binom{2k}{k}}x^{2k+1}\tag1 $$ Integrating $(1)$ and multiplying by $4$ yields $$ \begin{align} 2\arcsin^2(x) &=\sum_{k=0}^\infty\frac1{(2k+1)(2k+2)}\frac{4^{k+1}}{\binom{2k}{k}}x^{2k+2}\\ &=\sum_{k=0}^\infty\frac{4^{k+1}x^{2k+2}}{(k+1)^2\binom{2k+2}{k+1}}\\ &=\sum_{k=1}^\infty\frac{4^kx^{2k}}{k^2\binom{2k}{k}}\tag2 \end{align} $$ Applying $F_n=\frac{\phi^n-(1-\phi)^n}{\sqrt5}$, we get $$ \begin{align} \sum_{n=1}^\infty\frac{F_{2n}}{n^2\binom{2n}{n}} &=\frac2{\sqrt5}\left(\arcsin^2\left(\frac\phi2\right)-\arcsin^2\left(\frac{1-\phi}2\right)\right)\\ &=\frac2{\sqrt5}\left(\frac{9\pi^2}{100}-\frac{\pi^2}{100}\right)\\ &=\frac{4\pi^2}{25\sqrt5}\tag3 \end{align} $$

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Addendum: Trigonometric Values Used Above

Note that $$ \begin{align} 4\cos^2(2\pi/5)+2\cos(2\pi/5)-1 &=\left(e^{2\pi i/5}+e^{-2\pi i/5}\right)^2+\left(e^{2\pi i/5}+e^{-2\pi i/5}\right)-1\\[6pt] &=e^{4\pi i/5}+e^{2\pi i/5}+1+e^{-2\pi i/5}+e^{-4\pi i/5}\\ &=\frac{e^{10\pi i/5}-1}{e^{2\pi i/5}-1}\,e^{-4\pi i/5}\\ &=0\tag4 \end{align} $$ Solving for the positive root of $(4)$ gives $$ \begin{align} \cos(2\pi/5) &=\frac{-1+\sqrt{5}}4\\ &=\frac{\phi-1}2\tag5 \end{align} $$ Since $\cos(2\pi/5)=2\cos^2(\pi/5)-1$ and $\cos(\pi/5)$ is positive, apply $\phi^2=\phi+1$ to get $$ \cos(\pi/5)=\frac\phi2\tag6 $$ Since $\cos(x)=\sin(\pi/2-x)$, $(5)$ and $(6)$ become $$ \begin{align} \sin(\pi/10)&=\frac{\phi-1}2\tag7\\ \sin(3\pi/10)&=\frac\phi2\tag8 \end{align} $$