How to prove the identity $L(2,(\frac{\cdot}3))=\frac2{15}\sum\limits_{k=1}^\infty\frac{48^k}{k(2k-1)\binom{4k}{2k}\binom{2k}k}$?

Question

For the Dirichlet character $\chi(a)=(\frac a3)$ (which is the Legendre symbol), we have $$L(2,\chi)=\sum_{n=1}^\infty\frac{(\frac n3)}{n^2}=0.781302412896486296867187429624\ldots.$$ Note that this series converges slowly.

In 2014, motivated by my conjectural congruence $$\sum_{k=1}^{p-1}\frac{\binom{4k}{2k+1}\binom{2k}k}{48^k}\equiv\frac5{12}p^2B_{p-2}\left(\frac13\right)\pmod{p^3}\ \ \ \text{for any prime}\ p>3$$ (cf. Conjecture 1.1. of my paper available from http://maths.nju.edu.cn/~zwsun/165s.pdf), I found the following rapidly convergent series for the constant $L(2,(\frac{\cdot}3))$:

$$L\left(2,\left(\frac{\cdot}3\right)\right)=\frac2{15}\sum _{k=1}^\infty\frac{48^k}{k(2k-1)\binom{4k}{2k}\binom{2k}k}.\tag{1}$$ As the right-hand side of (1) converges quickly, you will not doubt the truth of (1) if you use Mathematica or Maple to check it. Unlike Ramanujan-type series for $1/\pi$, the summand in (1) just involves a product of two (not three) binomial coefficients. Note that $(1)$ was listed as $(1.9)$ in my preprint List of conjectural series for powers of $\pi$ and other constants.

QUESTION: How to prove my conjectural identity $(1)$?

I have mentioned this question to several experts at $\pi$-series or hypergeometric series, but none of them could prove the identity $(1)$. Any helpful ideas towards the proof of $(1)$?

Answer 1

This has been recently proved in my article (the last page). The idea is transparent enough to be outlined here.

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For $a,b$ near $0$, using Pochhammer symbol, we have the hypergeometric identity $$\tag{*}\small {\sum_{k\geq 0} \frac{2 \left(-\frac{1}{3}\right)^k \left(-a+b+\frac{1}{2}\right)_k (2 a+b+1)_k}{(4 a+2 b+2 k+1) (b+1)_k \left(2 a+b+\frac{1}{2}\right)_k} = \frac{\pi 4^a 3^{-a+b-\frac{1}{2}} \Gamma (b+1) \sec (\pi (a-b)) \Gamma \left(2 a+b+\frac{1}{2}\right)}{\Gamma (a+1) \Gamma \left(-a+b+\frac{1}{2}\right) \Gamma (2 a+b+1)} \\ - b\sum_{n\geq 1} \frac{3^n \left(a+\frac{1}{2}\right)_n \left((a+1)_n\right){}^2 (2 a+b+1)_{2 n}}{(a+n) (2 a+b+2 n-1) (2 a+1)_{2 n} \left(a-b+\frac{1}{2}\right)_n \left(2 a+b+\frac{1}{2}\right)_{2 n}} }$$ both sides are analytic in $a,b$, comparing coefficient of $a^0b^1$ gives $$\sum _{k=0}^{\infty } -\frac{4 \left(-\frac{1}{3}\right)^k}{(2 k+1)^2} = \frac{\pi \log (3)}{\sqrt{3}} - \sum _{n=1}^{\infty } \frac{3^n \left((1)_n\right){}^2}{n (2 n-1) \left(\frac{1}{2}\right)_{2 n}}$$

The sum on RHS is exactly OP's series. LHS equals $2\sqrt{3}i(\text{Li}_2(\frac{i}{{\sqrt 3 }}) - \text{Li}_2(\frac{{ - i}}{{\sqrt 3 }}))$, which can be shown, using functional equations of $\text{Li}_2$ or other methods, equals to $\frac{\pi \log (3)}{\sqrt{3}}-\frac{15 L(\chi,2)}{2}$, completing the proof when we assume $(*)$.

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The proof of $(*)$ is in style of WZ-pair. If $F(n,k), G(n,k)$ are two $\mathbb{C}$-valued functions satisfying $$\tag{1}F(n+1,k) - F(n,k) = G(n,k+1)-G(n,k)$$ then via some telescoping, one has, if $\lim_{n\to \infty} G(n,k) = 0$ for each $k\geq 0$, then $$\sum_{k\geq 0} F(0,k) = \lim_{n\to\infty} \sum_{k\geq 0} F(n,k) + \sum_{n\geq 0} G(0,n)$$ Now take $$F(n,k) = \frac{(-1)^k 3^{n-k} \Gamma (a+n+1)^2 \Gamma \left(-a+b+k-n+\frac{1}{2}\right) \Gamma (2 a+b+k+2 n+1)}{\Gamma \left(-a-n+\frac{1}{2}\right) \Gamma (2 a+2 n+1) \Gamma (b+k+1) \Gamma \left(2 a+b+k+2 n+\frac{3}{2}\right)}$$

one checks $$G(n,k) = \frac{3 (b+k) (2 a+b+k+2 n+2)}{(2 a-2 b-2 k+2 n+1) (4 a+2 b+2 k+4 n+3)} F(n,k)$$ satisfies $(1)$. With some rewriting, $\sum_{k\geq 0} F(0,k)$ gives LHS of $(*)$; $\sum_{n\geq 0} G(0,n)$ gives summation on RHS; some acrobatics in asymptotic analysis shows $\lim_{n\to\infty} \sum_{k\geq 0} F(n,k)$ gives the gamma product in $(*)$, proving $(*)$.

Answer 2

To provide an alternative approach, I'll cross-post my answer from MSE, adapted for the series.
Similarly to what @Nemo obtained in the comments, we can transform the original series into an equivalent integral, by utilizing the following identities:

$$\sqrt 3 \frac{\arcsin \left(\frac{\sqrt 3}{2} x\right)}{\sqrt{1-\frac{3}{4} x^2}} = \sum_{n=1}^\infty \frac{3^n x^{2n-1}}{n\binom{2n}{n}}$$

$$\int_0^\frac{\pi}{2}\left(4\sin^{4n-3}x-2\sin^{4n-1}x\right)dx=\frac{16^n}{(2n-1)\binom{4n}{2n}}$$

Furthermore, to evaluate the resulting integral we will use a long chain of substitutions aimed to arrive at a point where we can smoothly switch to double integrals by using:

$$ \frac{\arctan\left(f(x)\frac{\sqrt{g(x)}}{\sqrt{h(x)}} \right)}{\sqrt{g(x)}\sqrt{h(x)}}dx=\int_0^{f(x)} \frac{1}{h(x)+g(x)y^2}dy$$

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$$\sum_{n=1}^\infty \frac{48^n }{n(2n-1)\binom{2n}{n}\binom{4n}{2n}}=2\sqrt 3 \int_0^\frac{\pi}{2} \frac{2-\sin^2 x}{\sin x}\frac{\arcsin \left(\frac{\sqrt 3}{2} \sin^2 x\right)}{\sqrt{1-\frac{3}{4} \sin^4x}}dx$$

$$\overset{\sin^2 x\to \cos x}=\sqrt 6\int_0^\frac{\pi}{2}\frac{(2-\cos x)\cos \frac{x}{2}}{\cos x}\frac{\arcsin \left(\frac{\sqrt 3}{2} \cos x\right)}{\sqrt{1-\frac{3}{4} \cos^2x}}dx$$

$$\overset{\frac{\pi}{2}-x\to \arcsin x}=\sqrt 6\int_0^1 \frac{\color{blue}{(1-x)^{3/2}}+(1+x)^{3/2}}{(1-x^2)\sqrt{1+3x^2}}\arcsin\left(\frac{\sqrt 3\sqrt{1-x^2}}{2}\right)dx$$

$$\overset{{\color{blue}{x\to -x}}}=\sqrt 6\int_{-1}^1 \frac{(1+x)^{3/2} \arcsin\left(\small \frac{\sqrt 3\sqrt{1-x^2}}{2}\right)}{(1-x^2)\sqrt{1+3x^2}}dx \overset{x\to \frac{1-x}{1+x}} = \sqrt 3\int_0^\infty \frac{\arcsin\left(\frac{\sqrt{3x}}{1+x}\right)}{x\sqrt{1+x^3}}dx$$

$$=\sqrt{3}\int_0^\infty \frac{\arctan \left(\frac{\color{red}{\sqrt{3x(1+x)}}}{\sqrt{1+x^3}}\right)}{x\sqrt{1+x^3}}dx=\sqrt{3}\int_0^\infty \int_0^{\color{red}{f(x)}} \frac{1}{x(1+x^3+y^2)}dydx$$

$$\overset{y\to f(y)}=\sqrt{3} \int_0^\infty \int_0^x \frac{f'(y)}{x(1+x^3+f^2(y))}dydx=\int_0^\infty \int_0^x (*)dydx = \int_0^\infty \int_y^\infty (*)dxdy$$

$$ = \frac{1}{\sqrt 3}\int_0^\infty \left(\frac{\ln \left(\frac{x^3}{1+x^3+f^2(y)}\right)}{1+f^2(y)}\right) \bigg|_{x\to y}^{x\to \infty} dy = \frac{3}{2}\int_0^\infty \frac{(1+2y)\ln\left(\frac{1+y}{y}\right)}{\sqrt{y(1+y)}(1+3y+3y^2)}dy$$

$$\overset{\frac{y}{1+y}\to y^2}=-6 \int_0^1 \frac{(1+y^2)\ln y}{1+y^2+y^4}dy =2\sqrt{3}\sum_{n=1}^\infty \frac{\sin\left(\frac{n\pi}{3}\right)+\sin\left(\frac{2n\pi}{3}\right)}{n^2}$$

$$= 2\sqrt{3}\left(\operatorname{Cl}_2\left(\frac{\pi}{3}\right)+\operatorname{Cl}_2\left(\frac{2\pi}{3}\right)\right)=\frac{10}{\sqrt 3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right) = \boxed{\frac{15}{2} L(2,\chi)}$$

Where $\operatorname{Cl}_2(x)$ is the Clausen Function and $L(s,\chi)$ is the Dirichlet L-function.

Answer 3

Here is a related paper that shows a hypergeometric function similar to $R=\frac{1}{12}(-4 \ _4F_3(1,1,1,1;\frac{4}{3},\frac{3}{2},\frac{5}{3};-\frac{1}{4}) $ https://arxiv.org/pdf/1708.04269 Unfortunately, $R$ does not appear to be of beta-type even though it looks very similar to the one in the paper: $ ( \ _4F_3(1,1,1,\frac{3}{2};2,\frac{4}{3},\frac{5}{3};z) $ There is no way of making the $(3/2)_n$ pochammer part of the top go away and then replace it with the same in the bottom by performing integrations or taking derivatives of the function.