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Let $\phi$ be the famous golden ratio $\frac{\sqrt5+1}2$, and let $\zeta(3)=\sum_{n=1}^\infty\frac1{n^3}.$

In 2014, in the paper

Zhi-Wei Sun, New series for some special values of $L$-functions, Nanjing Univ. J. Math. Biquarterly 32 (2015), no. 2, pp 189–218, arXiv:1010.4298, author pdf.

I formulated the following conjecture involving both $\phi$ and $\zeta(3)$ (equations (3.11) and (3.12) in the above paper).

Conjecture. We have the identities $$\sum_{k=1}^\infty\frac{L_{2k}}{k^2\binom{2k}k}\left(\frac1k+\frac1{k+1}+\cdots+\frac1{2k}\right)=\frac{41\zeta(3)+4\pi^2\log\phi}{25}\tag{1}$$ and $$\sum_{k=1}^\infty\frac{v_k}{k^2\binom{2k}k}\left(\frac1k+\frac1{k+1}+\cdots+\frac1{2k}\right)=\frac{124\zeta(3)+\pi^2\log\left(5^5\phi^6\right)}{50}, \tag{2}$$ where the Lucas numbers $L_0,L_1,L_2,\ldots$ are given by $$L_0=2,\ L_1=1,\ \text{and}\ L_{n+1}=L_n+L_{n-1}\ \ \text{for all}\ n=1,2,3,\ldots,$$ and $v_0,v_1,v_2,\ldots,$ are defined by $$v_0=2,\ v_1=5,\ \text{and}\ v_{n+1}=5(v_n-v_{n-1})\ \ \text{for all}\ n=1,2,3,\ldots.$$

Remark. I found $(1)$ and $(2)$ on Nov. 29, 2014 and Dec. 7, 2014 respectively. As the two series converge rapidly, we can easily check $(1)$ and $(2)$ numerically. It is easy to show that $$\sum_{k=1}^\infty\frac{L_{2k}}{k^2\binom{2k}k}=\frac{\pi^2}5\ \ \text{and}\ \ \sum_{k=1}^\infty\frac{v_k}{k^2\binom{2k}k}=\frac25\pi^2.$$

QUESTION. How to prove the identities $(1)$ and $(2)$?

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  • $\begingroup$ Is this the source paper? maths.nju.edu.cn/~zwsun/153p.pdf $\endgroup$ Commented May 28, 2019 at 1:01
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    $\begingroup$ Not that one. The correct source paper is maths.nju.edu.cn/~zwsun/165s.pdf $\endgroup$ Commented May 28, 2019 at 1:38
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    $\begingroup$ Since $2(\arcsin x)^2=\sum_{n\ge 1}\frac{(2x)^{2n}}{n^2\binom{2n}{n}}$ and the integrals $\int \frac{(\arcsin (a x))^2}{x} \, dx$, $\int \frac{(\arcsin a)^2-(\arcsin (a x))^2}{1-x} \, dx$, $\int \frac{(\arcsin a)^2-(\arcsin (a \sqrt{x}))^2}{1-x} \, dx$ according to Mathematica have closed forms in terms of Polylogarithms, the sums $\sum_{n\ge 1}\frac{x^{2n}}{n^3\binom{2n}{n}}$, $\sum_{n\ge 1}\frac{x^{2n}}{n^2\binom{2n}{n}}H_n$,$\sum_{n\ge 1}\frac{x^{2n}}{n^2\binom{2n}{n}}H_{2n}$ also have closed form in terms of Polylogarithms. $\endgroup$
    – Nemo
    Commented May 28, 2019 at 7:56

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