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

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 $(1)$?

• In 2010 I conjectured that $$L\left(2,\left(\frac{\cdot}3\right)\right)=\sum_{k=1}^\infty\frac{(15k-4)(-27)^{k-1}}{k^3\binom{2k}k^2\binom{3k}k}$$ which was confirmed by Kh. Hessami Pilehrood and T. Hessami Pilehrood [Electron. J. Combin. 18(2012), #P35]. Using this, we can check (1) numerically. Jun 17, 2018 at 14:23
• Both sides should be periods and it might be possible to directly compare the motives and show they are isomorphic. Jun 17, 2018 at 14:32
• @Will Sawin: Indeed, Mathematica gives for the sum (1): $\frac{8}{15} \ _4F_3(\frac{1}{2},1,1,2;\frac{5}{4},\frac{3}{2},\frac{7}{4};\frac{3}{4})$. Jun 17, 2018 at 15:09
• The paper of Kh. & T. Hessami Pilehrood, cited in the comments above can be found here: combinatorics.org/ojs/index.php/eljc/article/view/v18i2p35 . In their notation, $K$ is the constant of interest in this question. The result is proved on page 10 after Corr. 4, using the following identity involving Hurwitz zeta functions: $9K = \zeta(2,1/3)-\zeta(2,2/3)$.
– j.c.
Jun 17, 2018 at 19:39
• Alternative form $$\int_0^{\pi/3}\frac{\left(2-\sqrt{3} \sin y\right) (y-\sin y\cos y)}{\sin ^3y \sqrt{3-2 \sqrt{3} \sin y}}dy=\frac{5}{4}L(2,\chi)$$
– Nemo
Jun 18, 2018 at 9:55

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 $$(*)$$.
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 $$(*)$$.