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Proof of conjectures on series with summands involving $ \binom{2k}{k}8^k/(\binom{3k}{k}\binom{6k}{3k})$

Zhi-Wei Sun, Yajun Zhou

Abstract

Using cyclotomic multiple zeta values of level $8$, we confirm and generalize several conjectural identities on infinite series with summands involving $\binom{2k}k8^k/(\binom{3k}k\binom{6k}{3k})$. For example, we prove that \[\sum_{k=0}^\infty\frac{(350k-17)\binom{2k}k8^k} {\binom{3k}k\binom{6k}{3k}}=15\sqrt2\,π+27\] and \[\sum_{k=1}^\infty\frac{\left\{(5k-1)\left[16\mathsf H_{2k-1}^{(2)}-3\mathsf H_{k-1}^{(2)}\right]-\frac{12(6k-1)}{(2k-1)^2}\right\}\binom{2k}k8^k} {k(2k-1)\binom{3k}k\binom{6k}{3k}}=\frac{π^3}{12\sqrt2},\] where $\mathsf H^{(2)}_m$ denotes the second-order harmonic number $\sum_{0<j\leq m}\frac1{j^2}$.

Proof of conjectures on series with summands involving $ \binom{2k}{k}8^k/(\binom{3k}{k}\binom{6k}{3k})$

Abstract

Using cyclotomic multiple zeta values of level , we confirm and generalize several conjectural identities on infinite series with summands involving . For example, we prove that and \[\sum_{k=1}^\infty\frac{\left\{(5k-1)\left[16\mathsf H_{2k-1}^{(2)}-3\mathsf H_{k-1}^{(2)}\right]-\frac{12(6k-1)}{(2k-1)^2}\right\}\binom{2k}k8^k} {k(2k-1)\binom{3k}k\binom{6k}{3k}}=\frac{π^3}{12\sqrt2},\] where denotes the second-order harmonic number .
Paper Structure (9 sections, 8 theorems, 39 equations, 6 tables)

This paper contains 9 sections, 8 theorems, 39 equations, 6 tables.

Key Result

Lemma 2.1

We have the following identities for $k\in\mathbb Z_{>0}$:

Theorems & Definitions (16)

  • Lemma 2.1
  • proof
  • Lemma 2.2
  • proof
  • Theorem 3.1
  • proof
  • Corollary 3.2
  • proof
  • Proposition 3.3
  • proof
  • ...and 6 more