含有中心二项式系数以及广义调和数的无穷级数恒等式

摘要/Abstract

摘要：

该文以两个高斯超几何求和公式为基础，建立一系列关于中心二项式系数和广义调和数的无穷级数恒等式.

Abstract:

In this paper, by appling Gauss's two hypergeometric summation formulas, we derive some infinite series expressions for the central binomial coefficients and the generalized harmonic numbers.

Key words: Hypergeometric series, Central binomial coefficients, Generalized harmonic numbers, Summation formulas

中图分类号:

O157.1

刘红梅. 含有中心二项式系数以及广义调和数的无穷级数恒等式[J]. 数学物理学报, 2020, 40(3): 631-640.

Hongmei Liu. Infinite Series Involving Central Binomial Coefficients and Generalized Harmonic Numbers[J]. Acta mathematica scientia,Series A, 2020, 40(3): 631-640.

图/表 3

表 1

由推论2.1给出的关于中心二项式系数的求和公式"

$m$	$\lambda$	$\mu$	$\sigma$	central binomial coefficient summation formulas
$0$				$\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(n+1)}=2$
$1$	$1$	$0$	$0$	$\sum\limits_{n=1}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(n+1)}H_n=4\ln2$ [13, (2.11)]
$1$	$0$	$1$	$0$	$\sum\limits_{n=1}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(n+1)}O_n=2$ [13, (2.13)]
$1$	$0$	$0$	$1$	$\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(n+1)}H_{n+1}=4$ [13, (2.12)]
$2$	$1$	$0$	$0$	$\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(n+1)}(H_n^2-H_n^{(2)})=8 \ln^22+\frac{2}{3}\pi^2$ [14, Ex.4.1]
$2$	$0$	$1$	$0$	$\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(n+1)}(O_n^2-O_n^{(2)})=4$[14, Ex.4.22]
$2$	$0$	$0$	$1$	$\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(n+1)}(H_{n+1}^{(2)}+H_{n+1}^2 -2H_{n+1})=8$ [14, Ex.4.24]
$3$	$1$	$0$	$0$	$\begin{array}{l}\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n})}{4^n(n+1)}(H_{n}^{3}-3H_{n}H_n^{(2)}+2H_{n}^{(3)})\\ =4(4\ln^32+\pi^2\ln2+6\zeta(3)) \ {\rm [14, \; Ex.4.11]} \end{array}$
$3$	$0$	$1$	$0$	$\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n})}{4^n(n+1)}(O_{n}^{3}-3O_{n}O_n^{(2)}+2O_{n}^{(3)})=12$[14, Ex.4.23]
$3$	$0$	$0$	$1$	$ \begin{array}{l}\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n})}{4^n(n+1)}(H_{n+1}^3-3H_{n+1}^2+3H_{n+1}H_{n+1}^{(2)} \\ -3H_{n+1}^{(2)}+2H_{n+1}^{(3)})=48\ {\rm [14, \; Ex.4.25]}\end{array}$
$4$	$1$	$0$	$0$	$\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(n+1)}(H_{n}^4-6H_{n}^2H_n^{(2)}+3(H_{n}^{(2)})^2+8H_{n}H_n^{(3)}-6H_{n}^{(4)})\\ =2(16\ln^42+8\pi^2\ln^22+\pi^4/3+96\ln2\cdot\zeta(3)+84\zeta(4)) \end{array}$
$4$	$0$	$1$	$0$	$\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(n+1)}(O_{n}^{4}-6O_{n}^2O_n^{(2)}+3(O_n^{(2)})^2+8O_nO_{n}^{(3)}-6O_n^{(4)})=48$
$4$	$0$	$0$	$1$	$\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n})}{4^n(n+1)}(-4H_{n+1}^3+H_{n+1}^4-12H_{n+1}H_{n+1}^{(2)}+6H_{n+1}^2H_{n+1}^{(2)} \\+3(H_{n+1}^{(2)})^2-8H_{n+1}^{(3)}+8H_{n+1}H_{n+1}^{(3)}+6H_{n+1}^{(4)})=384 \end{array}$

表 1

表 2

表 3

由推论2.3给出的关于中心二项式系数的求和公式"

$m$	$\lambda$	$\mu$	$\sigma$	central binomial coefficient summation formulas
$0$				$\sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n}) }^2}{16^n(n+1)}=\frac{4}{\pi}$
$1$	$1$	$0$	$0$	$\sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n}) }^2}{16^n(n+1)}O_n=\frac{4}{\pi}(1-\ln2)$[13, (2.17)]
$1$	$0$	$0$	$1$	$\sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n}) }^2}{16^n(n+1)}H_{n+1}=\frac{16}{\pi}(1-\ln2)$[13, (2.18)]
$2$	$1$	$0$	$0$	$\sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n})}^2}{16^n(n+1)}(O_n^2-O_n^{(2)}) =\frac{4}{\pi}(2-2\ln2+\ln^22-\frac{\pi^2}{12})$ [14, Ex.4.2]
$2$	$0$	$0$	$1$	$\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n})}^2}{16^n(n+1)}(H_{n+1}^{(2)}\!+\!H_{n+1}^2\!-\!2H_{n+1}) \!=\!\frac{4}{\pi}(16\ln^22\!-\!24\ln2\!+\!16\!-\!\frac{2\pi^2}{3}) {\rm [14, Ex.4.7]} \end{array}$
$3$	$1$	$0$	$0$	$\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n}) }^2}{16^n(n+1)}(O_{n}^3-3O_{n}O_n^{(2)}+2O_{n}^{(3)}) \\ =\frac{1}{\pi}(24-16\ln2+4\ln^22-4\ln^32-\pi^2+\pi^2\ln2-6\zeta(3))\ {\rm [14, Ex.4.12]} \end{array}$
$3$	$0$	$0$	$1$	$\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n}) }^2}{16^n(n+1)}(H_{n+1}^3 -3H_{n+1}^2+3H_{n+1}H_{n+1}^{(2)}-3H_{n+1}^{(2)}+2H_{n+1}^{(3)}) \\ =\frac{8}{\pi}(60-60\ln2+24\ln^22-32\ln^32-3\pi^2+4\pi^2\ln2-12\zeta(3))\ {\rm [14, Ex.4.13]} \end{array}$
$4$	$1$	$0$	$0$	$\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n})}^2}{16^n(n+1)}(O_{n}^{4}-6O_{n}^2O_n^{(2)} +3(O_n^{(2)})^2+8O_nO_{n}^{(3)}-6O_n^{(4)})\\ =\frac{4}{\pi}(24-24\ln2+12\ln^22-4\ln^32+\ln^42-\pi^2+\pi^2\ln2 -\frac{1}{2}\pi^2\ln^22\\ +\frac{\pi^4}{48}-6\zeta(3)+6\zeta(3)\ln2-\frac{21}{4}\zeta(4)) \end{array}$
$4$	$0$	$0$	$1$	$\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{{({}^{2n}_{\; n}) }^2}{16^n(n+1)}(-4H_{n+1}^3+H_{n+1}^4-12H_{n+1}H_{n+1}^{(2)} +6H_{n+1}^2H_{n+1}^{(2)}\\ +3(H_{n+1}^{(2)})^2-8H_{n+1}^{(3)}+8H_{n+1}H_{n+1}^{(3)}+6H_{n+1}^{(4)}) \\ =\frac{4}{\pi}(1152-1920\ln2+1536\ln^22-768\ln^32+256\ln^42-64\pi^2+96\pi^2\ln2 \\ -64\pi^2\ln^22+\frac{4}{3}\pi^4-288\zeta(3)+384\ln2\cdot\zeta(3)-168\zeta(4)) \end{array}$

表 3

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