含有中心二项式系数以及广义调和数的无穷级数恒等式
刘红梅

Infinite Series Involving Central Binomial Coefficients and Generalized Harmonic Numbers
Hongmei Liu
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}$
$4$ $1$ $0$ $0$ $\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(n+1)}(H_{n}^4-6H_{n}^2H_n^{(2)}+3(H_{n}^{(2)})^2+8H_{n}H_n^{(3)}-6H_{n}^{(4)})\\ =2(16\ln^42+8\pi^2\ln^22+\pi^4/3+96\ln2\cdot\zeta(3)+84\zeta(4)) \end{array}$
$4$ $0$ $1$ $0$ $\sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n}) }{4^n(n+1)}(O_{n}^{4}-6O_{n}^2O_n^{(2)}+3(O_n^{(2)})^2+8O_nO_{n}^{(3)}-6O_n^{(4)})=48$
$4$ $0$ $0$ $1$ $\begin{array}{l} \sum\limits_{n=0}^{\infty}\frac{({}^{2n}_{\; n})}{4^n(n+1)}(-4H_{n+1}^3+H_{n+1}^4-12H_{n+1}H_{n+1}^{(2)}+6H_{n+1}^2H_{n+1}^{(2)} \\+3(H_{n+1}^{(2)})^2-8H_{n+1}^{(3)}+8H_{n+1}H_{n+1}^{(3)}+6H_{n+1}^{(4)})=384 \end{array}$