8
$\begingroup$

Recently, I found a (conjectural) new series for $\sqrt3\pi$: $$\sum_{k=1}^\infty\frac{(8k-3)\binom{4k}{2k}}{k(4k-1)9^k\binom{2k}k^2}=\frac{\sqrt3\pi}{18}.\label{1}\tag{1}$$ The series converges fast with converging rate $1/9$, so one can easily check the identity \eqref{1} numerically.

Motivated by \eqref{1}, I also conjecture that $$\sum_{k=1}^\infty\frac{\binom{4k}{2k}\left((8k-3)(5H_{2k-1}-4H_{k-1})-6\right)}{k(4k-1)9^k\binom{2k}k^2}=\frac32\sum_{n=0}^\infty\left(\frac1{(3n+1)^2}-\frac1{(3n+2)^2}\right) \label{2}\tag{2}$$ and $$\sum_{k=1}^\infty\frac{\binom{4k}{2k}(8k-3)\left(2H_{2k-1}^{(2)}-5H_{k-1}^{(2)}\right)}{k(4k-1)9^k\binom{2k}k^2}=\frac{\pi^3}{36\sqrt3}, \label{3}\tag{3}$$ where $$H_n:=\sum_{0<k\le n}\frac1k\ \ \ \text{and}\ \ \ \ H_n^{(2)}:=\sum_{0<k\le n}\frac1{k^2}$$ for each $n=0,1,2,\ldots$.

QUESTION. Can one prove the new identities \eqref{1},\eqref{2} and \eqref{3} by using known tools (such as the WZ method and various hypergeometric series identities) ?

For the identities \eqref{1},\eqref{2} and \eqref{3}, we also have corresponding conjectural $p$-adic congruences. For example, I conjecture that for any prime $p>3$ we have $$\sum_{k=1}^{(p-1)/2}\frac{(8k-3)\binom{4k}{2k}}{k(4k-1)9^k\binom{2k}k^2}\equiv-\frac5{36}pB_{p-2}\left(\frac13\right)\pmod{p^2}\tag{4}$$ and $$\sum_{k=1}^{(p-1)/2}\frac{\binom{4k}{2k}\left((8k-3)(5H_{2k-1}-4H_{k-1})-6\right)}{k(4k-1)9^k\binom{2k}k^2}\equiv\frac16B_{p-2}\left(\frac13\right)\pmod p,\tag{5}$$ where $B_{p-2}(x)$ denotes the Bernoulli polynomial of degree $p-2$.

Your comments are welcome!

$\endgroup$
9
  • 4
    $\begingroup$ How did you find this?? $\endgroup$ Commented Oct 20, 2023 at 1:47
  • 2
    $\begingroup$ Equation (2) seems to have the closed form $\frac{2\,\sqrt{3}\,Gi}{3}$, where $Gi$ is Gieseking's constant ( mathworld.wolfram.com/GiesekingsConstant.html ). $\endgroup$
    – Agno
    Commented Oct 20, 2023 at 12:17
  • 1
    $\begingroup$ @SidharthGhoshal I don't know how the author found these identities, but I can suggest a possible way using a computer: The first identity is an identity for a generalized hypergeometric series $_5F_4$; namely, it states that $_5F_4(3/4,1,1,5/4,13/8; 5/8,3/2,3/2,3/2; 1/9)=\pi\sqrt3/5$. Also, the parametric excess of a hypergeometric is the sum of its bottom parameters, minus the sum of its top parameters, and is an important quantity in the study of hypergeometric series, and it happens to be $-1/2$ in this case… $\endgroup$ Commented Oct 21, 2023 at 11:09
  • 1
    $\begingroup$ … So, you could compute many $_5F_4$ values with integer or half-integer parametric excess at various rational entries with small denominators, perhaps out to a hundred decimal places or so, then you could use an inverse symbolic calculator to guess a nice closed form in terms of common irrational numbers such as square roots of natural numbers, $e$, and $\pi$. I think in the literature on experimental mathematics there should be discussions of exhaustive searches of certain input values for interesting hypergeometric identities. $\endgroup$ Commented Oct 21, 2023 at 11:09
  • 1
    $\begingroup$ Finally, I note that there is a lot of sophisticated hypergeometric-identity proving going on at Math StackExchange, with many interesting methods presented; see for example math.stackexchange.com/q/3736208 and math.stackexchange.com/q/2123298 and related posts. $\endgroup$ Commented Oct 21, 2023 at 11:09

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.