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Can anyone help me to prove the following relation.

$$\sum_{k=1}^{\infty} \frac{F_{2k}H^{(2)}_{k-1}}{k^2\binom{2k}{k}}=\frac{2\pi^4}{375\sqrt{5}}$$ I was studying recently about Fibonacci and Lucas numbers.

And I came through the above relationship. I tried applying golden ratio but nothing works. Symbols have their usual meanings.

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  • $\begingroup$ This is nice! Perhaps someone can answer it !! :) $\endgroup$ – Tolaso Oct 17 '16 at 8:08
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    $\begingroup$ Let me bump this.. $\endgroup$ – Tolaso Jan 5 '17 at 13:08
  • $\begingroup$ Where did you read the stuff, that should be a nice read! could you please share? $\endgroup$ – Math-fun May 29 '17 at 9:30
  • $\begingroup$ Since it is an older post, I didn't remember exactly. I have read it in research papers regarding Fibonacci and lucas numbers $\endgroup$ – Aman Rajput May 29 '17 at 10:15
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It is well known that $$\sum_{k=1}^{\infty}\frac{x^{2k}}{k^{2}\dbinom{2k}{k}}H_{k-1}^{\left(2\right)}=\frac{2\arcsin^{4}\left(\frac{x}{2}\right)}{3},\,\left|x\right|\leq2$$ (see here or here for a proof), then from the Binet formula we get $$\sum_{k=1}^{\infty}\frac{F_{2n}}{k^{2}\dbinom{2k}{k}}H_{k-1}^{\left(2\right)}=\frac{1}{\sqrt{5}}\sum_{k=1}^{\infty}\frac{\left(1+\sqrt{5}\right)^{2n}}{2^{2n}k^{2}\dbinom{2k}{k}}H_{k-1}^{\left(2\right)}-\frac{1}{\sqrt{5}}\sum_{k=1}^{\infty}\frac{\left(1-\sqrt{5}\right)^{2n}}{2^{2n}k^{2}\dbinom{2k}{k}}H_{k-1}^{\left(2\right)}$$ $$=\frac{2\arcsin^{4}\left(\frac{1+\sqrt{5}}{4}\right)}{3\sqrt{5}}-\frac{2\arcsin^{4}\left(\frac{1-\sqrt{5}}{4}\right)}{3\sqrt{5}}=\color{red}{\frac{2\pi^{4}}{375\sqrt{5}}}.$$

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