Recently, I found the following three (conjectural) identities for $\pi^2$: $$\sum_{k=1}^\infty\frac{145k^2-104k+18}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{\pi^2}3,\tag{1}$$ $$\sum_{k=1}^\infty\frac{(42k^2-23k+3)16^k}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{\pi^2}2,\tag{2}$$ $$\sum_{k=1}^\infty\frac{(92k^2-61k+9)64^k}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=8\pi^2.\tag{3}$$ The converging rates of the series in $(1)$-$(3)$ are $4/729$, $1/64$ and $4/27$, respectively. One can easily check the identities $(1)$-$(3)$ numerically.

Let $P(k)=145k^2-104k+18$. Motivated by $(1)$, I also conjecture the following identities involving harmonic numbers:

$$\sum_{k=1}^\infty\frac{6P(k)(H_{3k-1}-H_{k-1})-232k+89}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=18\zeta(3),\tag{4}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{2k-1}-H_{k-1})-\frac{3(58k^2-40k+7)}{2(2k-1)}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\zeta(3),\tag{5}$$ $$\sum_{k=1}^\infty\frac{P(k)(H_{3k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{17k+32}{9k}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{7\pi^4}{180},\tag{6}$$ $$\sum_{k=1}^\infty\frac{P(k)(297H_{3k-1}^{(2)}-192H_{2k-1}^{(2)}-978H_{k-1}^{(2)})+\frac{27(180k^2+12k-35)}{(2k-1)^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{167}{20}\pi^4,\tag{7}$$ and $$\sum_{k=1}^\infty\frac{P(k)H_{k-1}^{(3)}+\frac{28(2k-1)}{17k^2}}{k^3(2k-1)\binom{2k}k\binom{3k}k^2}=\frac{528\zeta(5)-46\pi^2\zeta(3)}{17},\tag{8}$$ where $H_n:=\sum_{0<j\le n}\frac1j$ and $H_n^{(m)}:=\sum_{0<j\le n}\frac1{j^m}$ for $m=2,3,\ldots$.

Let $Q(k)=(3k-1)(14k-3)=42k^2-23k+3$. Motivated by $(2)$, I conjecture the following identities: $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(H_{2k-1}-H_{k-1})-\frac{196k^2-100k+13}{6(2k-1)}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{2\pi^2\log2-7\zeta(3)}6,\tag{9}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(H_{4k-1}-H_{2k-1})-\frac{28k^2-76k+19}{12(2k-1)}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{2\pi^2\log2+35\zeta(3)}{12},\tag{10}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(4H_{4k-1}^{(2)}-H_{2k-1}^{(2)}-2H_{k-1}^{(2)})-\frac{6k+1}{k}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{5\pi^4}{24},\tag{11}$$ $$\sum_{k=1}^\infty\frac{16^k\left(Q(k)(4H_{4k-1}^{(2)}-5H_{2k-1}^{(2)}-3H_{k-1}^{(2)})+\frac{32k(3k-1)}{(2k-1)^2}\right)}{k^3(2k-1)\binom{2k}k\binom{4k}{2k}^2}=\frac{\pi^4}{6}.\tag{12}$$

Let $R(k)=92k^2-61k+9$. Inspired by $(3)$, I also conjecture the following identities: $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(3H_{3k-1}-5H_{k-1})-60k+21\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=48\pi^2\log2-56\zeta(3),\tag{13}$$ $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(3H_{2k-1}-2H_{k-1})-\frac{124k^2-79k+13}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=112\zeta(3),\tag{14}$$ and $$\sum_{k=1}^\infty\frac{64^k\left(R(k)(6H_{4k-1}-H_{k-1})-\frac{344k^2-281k+59}{2k-1}\right)}{k^3(2k-1)\binom{2k}k\binom{3k}k\binom{4k}{2k}}=560\zeta(3).\tag{15}$$

QUESTION. Can one prove the new identities $(1)-(15)$ via known methods (including the WZ method and hypergeometric series identies)?

Your comments are welcome!


1 Answer 1


All identities except possibly $(8)$ can be proved elegantly using Wilf-Zeilberger pairs.

To be precise, if $F(n,k), G(n,k)$ are two $\mathbb{C}$-valued functions satisfying $$\tag{*}F(n+1,k) - F(n,k) = G(n,k+1)-G(n,k)$$ then via some telescoping, one has, if $\lim_{n\to \infty} G(n,k) = 0$ for each $k\geq 0$, then $$\sum_{k\geq 0} F(0,k) = \lim_{n\to\infty} \sum_{k\geq 0} F(n,k) + \sum_{n\geq 0} G(0,n)$$

If one takes $$F(n,k) = \frac{\Gamma (-a+c+n+1) \Gamma \left(a+k+n+\frac{1}{2}\right) \Gamma (-b+c+n+1) \Gamma \left(b+k+n+\frac{1}{2}\right)}{\Gamma (a-n) \Gamma (b-n) \Gamma \left(k+2 n+\frac{3}{2}\right) \Gamma \left(c+k+2 n+\frac{3}{2}\right) \Gamma (-a-b+c+2 n+1)}$$ then one finds its WZ-mate $G(n,k)$ (here $\lim_{n\to\infty}\sum_{k\geq 0}F(n,k) = 0)$, we arrive at $$\sum_{k\geq 0}\frac{32 \left(a+d+\frac{1}{2}\right)_k \left(b+d+\frac{1}{2}\right)_k}{(2 d+2 k+1) (2 c+2 d+2 k+1) \left(d+\frac{1}{2}\right)_k \left(c+d+\frac{1}{2}\right)_k} \\ = \sum_{n\geq 1}\frac{(1-a)_{n-1} (1-b)_{n-1} (-a+c+1)_{n-1} \left(a+d+\frac{1}{2}\right)_{n-1} (-b+c+1)_{n-1} \left(b+d+\frac{1}{2}\right)_{n-1} \times P}{\left(d+\frac{1}{2}\right)_{2 n} \left(c+d+\frac{1}{2}\right)_{2 n} (-a-b+c+1)_{2 n}}$$

here $P$ is a long polynomial that pops up in computing WZ-mate $G(n,k)$: $8 a^2 b^2-8 a^2 b c-16 a^2 b n+8 a^2 c n+8 a^2 n^2-8 a b^2 c-16 a b^2 n+8 a b c^2+8 a b c d+48 a b c n-4 a b c+8 a b d^2+32 a b d n-8 a b d+64 a b n^2-16 a b n+2 a b-8 a c^2 d-24 a c^2 n+4 a c^2-16 a c d^2-72 a c d n+24 a c d-104 a c n^2+52 a c n-8 a c-8 a d^3-56 a d^2 n+20 a d^2-128 a d n^2+88 a d n-14 a d-112 a n^3+96 a n^2-30 a n+3 a+8 b^2 c n+8 b^2 n^2-8 b c^2 d-24 b c^2 n+4 b c^2-16 b c d^2-72 b c d n+24 b c d-104 b c n^2+52 b c n-8 b c-8 b d^3-56 b d^2 n+20 b d^2-128 b d n^2+88 b d n-14 b d-112 b n^3+96 b n^2-30 b n+3 b+8 c^3 d+16 c^3 n-4 c^3+16 c^2 d^2+80 c^2 d n-24 c^2 d+104 c^2 n^2-56 c^2 n+8 c^2+8 c d^3+80 c d^2 n-20 c d^2+232 c d n^2-128 c d n+14 c d+224 c n^3-180 c n^2+44 c n-3 c+16 d^3 n+104 d^2 n^2-40 d^2 n+224 d n^3-168 d n^2+28 d n+168 n^4-176 n^3+58 n^2-6 n$

When $(a,b,c,d) = (0,0,0,0)$, above equality reduces to $$\frac{\pi^2}{2} = \sum_{k\geq 0} \frac{4}{(1+2k)^2} = \sum_{n\geq 0} 64^{-n} \frac{(1/2)_n (1)_n^3}{(1/4)_n^2 (3/4)_n^2} \frac{3-23n+42n^2}{n^3(2n-1)}$$ which is exactly $(2)$. Those involving harmonic number $(9),(10),(11),(12)$ can be proved by comparing coefficient in Taylor expansion of above equality around $(a,b,c,d) = (0,0,0,0)$, coefficients of $\sum_{k\geq 0}$ will give the so-called multiple $t$-values (special case of alternating multiple zeta values), whose values, when weight is small, are well-understood.

For $(1)$ and $(4)$-$(7)$, one repeats the above process, except now using another $F(n,k)$: $F(n,k) = \frac{\Gamma (-a+c+2 n+1) \Gamma (a+k+n+1) \Gamma (-b+c+2 n+1) \Gamma (b+k+n+1)}{\Gamma (a-n) \Gamma (b-n) \Gamma (k+2 n+2) \Gamma (c+k+3 n+2) \Gamma (-a-b+c+3 n+1)}$

Unfortunately, one cannot deduce $(8)$ in this way.

For $(3)$ and $(13)$-$(15)$, one repeats the above process, with $F(n,k)$: $F(n,k) = \frac{\Gamma (-a+c+n+1) \Gamma (a+k+2 n+1) \Gamma (-b+c+n+1) \Gamma (b+k+2 n+1)}{\Gamma \left(a+\frac{1}{2}\right) \Gamma \left(b+\frac{1}{2}\right) \Gamma \left(k+2 n+\frac{3}{2}\right) \Gamma (c+k+3 n+2) \Gamma \left(-a-b+c+n+\frac{1}{2}\right)}$


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