# A vanishing sum and related $p$-adic congruences

Recently I had a curious discovery. Namely, I have made the following conjectures.

Conjecture 1. We have the identity $$\sum_{k=0}^\infty\frac{(10k-1)\binom{3k}k\binom{6k}{3k}}{(2k+1)512^k}=0.\label{1}\tag{1}$$

Conjecture 2. For any odd prime $$p$$ and positive integer $$n$$, we have $$\frac1{(pn)^2}\bigg(\sum_{k=0}^{pn-1}\frac{(10k-1)\binom{3k}k\binom{6k}{3k}}{(2k+1)512^k}-\left(\frac{-2}p\right)\sum_{k=0}^{n-1}\frac{(10k-1)\binom{3k}k\binom{6k}{3k}}{(2k+1)512^k}\bigg)\in\mathbb Z_p,\label{2}\tag{2}$$ where $$(\frac{\cdot}p)$$ is the Legendre symbol and $$\mathbb Z_p$$ is the ring of $$p$$-adic integers.

Conjecture 3. Let $$p$$ be an odd prime. Then $$\sum_{k=0}^{(p-1)/2}\frac{(10k-1)\binom{3k}k\binom{6k}{3k}}{(2k+1)512^k} \equiv\left(\frac{-2}p\right)\bigg(\frac98p^2q_p(2)^2-\frac32p\,q_p(2)-1\bigg)\ \ (\text{mod}\ p^3),\label{3}\tag{3}$$ where $$q_p(2)$$ denotes the Fermat quotient $$(2^{p-1}-1)/p$$. Also, $$\sum_{k=0}^{p-1}\frac{(10k-1)\binom{3k}k\binom{6k}{3k}}{(2k+1)512^k} \equiv-\left(\frac{-2}p\right)+\frac{15}{16}p^2E_{p-3}\left(\frac14\right) \ \ (\text{mod}\ p^3),\label{4}\tag{4}$$ where $$E_{p-3}(x)$$ denotes the Euler polynomial of degree $$p-3$$.

Note that the series in \eqref{1} has converging rate $$27/32$$. All the three conjectures can be easily checked numerically.

QUESTION。 How to prove the identity \eqref{1} and related $$p$$-adic congruences \eqref{2}, \eqref{3} and \eqref{4}? Is the WZ method helpful to solve the three conjectures?

The Archimedean version $$(1)$$ can be somehow systematically proved using WZ-method.

We recall the following analytic fact (c.f. Propositon 2.1): if $$F,G$$ satisfies $$F(n+1,k) - F(n,k) = G(n,k+1)-G(n,k)$$, suppose

• $$\lim_{n\to\infty} G(n,k) = 0$$ for each $$k\geq 0$$
• $$\sum_{k\geq 0} F(0,k)$$ converges
• $$\sum_{n\geq 0} G(n,0)$$ converges

then $$\lim_{n\to\infty} \sum_{k\geq 0} F(n,k)$$ exists and is finite, also $$\tag{1}\sum_{k\geq 0} F(0,k) = \sum_{n\geq 0} G(n,0) + \lim_{n\to\infty} \sum_{k\geq 0} F(n,k)$$

In most applications of WZ-pair to infinite series, the term $$\lim_{n\to\infty} \sum_{k\geq 0} F(n,k)$$ is zero. But for this example, it will not.

Now let $$F(n,k)= \frac{2^{-2 k-3 n} \Gamma \left(-a+c-n+\frac{1}{2}\right) \Gamma \left(a+c+2 d+2 k+3 n+\frac{1}{2}\right)}{\Gamma \left(a+c-n+\frac{1}{2}\right) \Gamma (d+k+2 n+1) \Gamma (c+d+k+n+1)}$$ one computes, using Gopser's algorithm, its WZ-mate $$G$$, and checks above three bullets are satisfied when $$a,c,d$$ are in a neighborhood of $$0$$ with $$\Re(a)<0$$.

$$(1)$$ becomes, after some rearrangements, $$\tag{2}\sum_{k\geq 0} \frac{2^{-2 k} \left(a+c+2 d+\frac{1}{2}\right)_{2 k}}{(d+1)_k (c+d+1)_k} = L -\frac{1}{4}\times \sum_{n\geq 0} \frac{2^{-3 n} P\times (-a-c+\frac{1}{2})_n (a+c+2 d+\frac{1}{2})_{3 n}}{(d+2 n+1) (2 a-2 c+2 n+1) (d+1)_{2 n} \left(a-c+\frac{1}{2}\right)_n (c+d+1)_n}$$ here $$\small{P = 4 a^2+8 a c+8 a d+8 a n+4 c^2+24 c d+40 c n+16 c+16 d^2+40 d n+12 d+20 n^2+8 n-1}$$ and $$L = \frac{\Gamma(a+c+1/2)\Gamma(d+1)\Gamma(c+d+1)}{\Gamma(a+c+2d+1/2) \Gamma(1/2-a+c)} \times \lim_{n\to\infty} \sum_{k\geq 0} \frac{2^{-2 k-3 n}\Gamma(-a+c-n+\frac{1}{2}) \Gamma(a+c+2 d+2 k+3 n+\frac{1}{2})}{\Gamma \left(-a+c+\frac{1}{2}\right)\Gamma \left(a+c-n+\frac{1}{2}\right) \Gamma (d+k+2 n+1) \Gamma (c+d+k+n+1)}$$ We know this limit $$L$$ exists when $$a,c,d$$ are near $$0$$ and $$\Re(a)<0$$, but we cannot calculate $$L$$ naively using dominated convergence or anything that tries to move limit inside summation. More careful analysis (e.g. Laplace method) will nonetheless give $$L = 2^{-a+c+2 d-1/2} \frac{\Gamma (-a) \Gamma (d+1) \Gamma \left(a-c+\frac{1}{2}\right) \Gamma (c+d+1)}{\sqrt{\pi } \Gamma \left(-a-c+\frac{1}{2}\right) \Gamma \left(a+c+2 d+\frac{1}{2}\right)}$$

When $$c+d=0$$, LHS of $$(2)$$ can be summed using Gauss $$_2F_1$$, it can then be seen to equal to $$L$$. Therefore for $$a,c\in \mathbb{C}$$, $$\sum_{n\geq 0}\frac{2^{-3 n} \left(4 a^2+8 a n-4 c^2+4 c+20 n^2+8 n-1\right) \left(-a-c+\frac{1}{2}\right)_n \left(a-c+\frac{1}{2}\right)_{3 n}}{(-c+2 n+1)(2 a-2 c+2 n+1) (1)_n (1-c)_{2 n} \left(a-c+\frac{1}{2}\right)_n} = 0$$

OP's $$(1)$$ is above with $$a=c=0$$.