An interesting problem is to prove that: $$ \sum_{n=1}^\infty \frac{F_{2n}}{n^2 \binom{2n}{n}}=\frac{4\pi^2}{25 \sqrt 5}. $$
I know the proof, which uses the fact that $\dfrac{1}{n\binom{2n}{n}}=\int\limits_{0}^{1}t^n(1-t)^{n-1}dt$, but my question is: is there some really elementary way to prove it?
$$\begin{eqnarray*}\sum_{n\geq 1}\frac{x^n}{n^2 \binom{2n}{n}}&=&\sum_{n\geq 1}\frac{(n-1)!^2}{(2n)!} x^n = \sum_{n\geq 1}B(n,n)\frac{x^n}{2n}=\int_{0}^{1}\sum_{n\geq 1}y^{n-1}(1-y)^{n-1}\frac{x^n}{2n}\\&=&-\frac{1}{2}\int_{0}^{1}\frac{\log(1-xy+xy^2)}{y(1-y)}\,dy\end{eqnarray*} $$ hence by taking $\sigma_+=\frac{3+\sqrt{5}}{2}$ and $\sigma_-=\frac{3-\sqrt{5}}{2}$:
$$\begin{eqnarray*}\sum_{n\geq 1}\frac{F_{2n}}{n^2\binom{2n}{n}}&=&-\frac{1}{2\sqrt{5}}\int_{0}^{1}\frac{\log(1-\sigma_{+}(y-y^2))-\log(1-\sigma_{-}(y-y^2))}{y(1-y)}\,dy\\&=&-\frac{2}{\sqrt{5}}\int_{0}^{1}\frac{\log(1-\sigma_+(1-u^2)/4)-\log(1-\sigma_-(1-u^2)/4)}{1-u^2}\,du\end{eqnarray*}$$ and the last integral can be computed by exploiting the well-known relations between the dilogarithm and the golden ratio. Another technique is shown in one of the answers I am most proud of, a cooperative work by me, achille hui and Bhenni Benghorbal. It gives that the problem boils down to computing a squared arcsine.
Hint: In general, $~\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2\displaystyle{2n\choose n}}~=~2\arcsin^2x.~$ Use this in conjunction with Binet's formula for Fibonacci numbers.
