Fibonacci and Harmonic

Question

Here is a cute series that I came across and cannot seem to tackle at all.

Question: Calculate

$$\mathcal{S} = \sum_{n=1}^{\infty} \frac{F_n \mathcal{H}_{n-1}^{(2)}}{n^2 \binom{2n}{n}}$$

where $F_n$ denotes the $n$ -th Fibonacci number and $\mathcal{H}_n^{(s)}$ denotes the $n$ - th harmonic number of order $s$.

What techniques would one use to evaluate it? Should we invoke the Beta function for the binomial? I don't have any ideas on the problem!!

The answer is claimed to be $\displaystyle \frac{2\pi ^{4}}{375\sqrt{5}}$. I have not checked numerically though!

Answer 1

Hint. One may just use Ramanujan's identity $$ \arcsin^4\left(\frac x2\right)=\frac32 \sum_{n=2}^\infty \left\{\sum_{k=1}^{n-1}\frac1{k^2}\right\}\frac{x^{2n}}{n^2\binom{2n}{n}}, \qquad |x|\le2, \tag1 $$ proved by Berndt (Ramanujan’s Notebooks Part 1 p. 263) then recalling that $$ F_n=\frac{\varphi^n-(1-\varphi)^{n}}{\sqrt 5},\qquad \varphi=\frac{1+\sqrt{5}}2, $$ one gets

$$ \mathcal{S} = \sum_{n=1}^{\infty} \frac{F_n \mathcal{H}_{n-1}^{(2)}}{n^2 \binom{2n}{n}}=\frac{2\arcsin^4\left(\dfrac {\sqrt{\varphi}}2\right)-2\arcsin^4\left(\dfrac {\sqrt{1-\varphi}}2\right)}{3\sqrt 5} \tag2 $$

giving $\mathcal{S}\ne\dfrac{2\pi^{4}}{375\sqrt{5}}$, we have $$ \mathcal{S} = 0.06085736037597791544248567 \cdots $$ and$$ \dfrac{2\pi^{4}}{375\sqrt{5}} = 0.2323342391237277 \cdots. $$