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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 $(1)$?

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    $\begingroup$ 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. $\endgroup$ – Zhi-Wei Sun Jun 17 '18 at 14:23
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    $\begingroup$ Both sides should be periods and it might be possible to directly compare the motives and show they are isomorphic. $\endgroup$ – Will Sawin Jun 17 '18 at 14:32
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    $\begingroup$ @Will Sawin: Indeed, Mathematica gives for the sum (1): $\frac{8}{15} \ _4F_3(\frac{1}{2},1,1,2;\frac{5}{4},\frac{3}{2},\frac{7}{4};\frac{3}{4})$. $\endgroup$ – Johannes Trost Jun 17 '18 at 15:09
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    $\begingroup$ 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: $9K = \zeta(2,1/3)-\zeta(2,2/3)$. $\endgroup$ – j.c. Jun 17 '18 at 19:39
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    $\begingroup$ Alternative form $$ \int_0^{\pi/3}\frac{\left(2-\sqrt{3} \sin y\right) (y-\sin y\cos y)}{\sin ^3y \sqrt{3-2 \sqrt{3} \sin y}}dy=\frac{5}{4}L(2,\chi) $$ $\endgroup$ – user82588 Jun 18 '18 at 9:55

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