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 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 $(1)$?

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 $(1)$?

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 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 $(1)$ is listed as $(1.9)$ in my paper 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 $(1)$?

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 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 $(1)$?

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 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 be soon convinced that (1) does hold if you use Mathematica or Maple. Unlike Ramanujan-type serie for $1/\pi$, the summand in (1) just involves a product of two (not three) binomial coefficients. Note that $(1)$ is listed as $(1.9)$ in my paper 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 $(1)$?

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 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 $(1)$ is listed as $(1.9)$ in my paper 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 $(1)$?