0 引言

在文献[1]中, Ramanujan共提出了17个关于$1/\pi$的级数. 1987年之前, 仍然没有数学家可以真正处理这些级数. Borwein兄弟在文献[2]中第一次给出了Ramanujan级数的严格证明并得到了许多新的关于$1/\pi$的Ramanujan型级数. Chudnovsky兄弟则在文献[3]中给出了这些级数的一些拓展公式.

目前, 有关$1/\pi$的一些新的Ramanujan型级数已经被许多数学家研究并取得了大量的成果, 具体可参考文献[4-19].在文献[20]的基础上, 本文通过对Dougall $_5F_4$求和公式中所包含的参数进行求导运算来得到一些新的有关$1/\pi$的Ramanujan型级数.同时, 文章中还得到了许多新的关于其他常数的Ramanujan型级数.

本文需要下列有关特殊函数的一些基本概念.

定义0.1^[21] 对任意的复数$z\not= 0, -1, -2, \cdots$, 伽马函数$\Gamma(z)$可以定义为

$ \Gamma (z) = \mathop {\lim }\limits_{k \to \infty } \frac{{k!{k^{z-1}}}}{{{{(z)}_k}}}, $

(1)

其中$(z)_k$为Pochhammer符号.

定义0.2 若$z$为复数且$n$为自然数, 则通常的Pochhammer符号被定义为

$ {(z)_n} = z(z + 1)(z + 2) \cdots (z + n-1), \quad {(z)_0} = 1. $

一般地, 对于任意的复数$\alpha$, 可定义Pochhammer符号$(z)_{\alpha}$为

$ {(z)_\alpha } = \frac{{\Gamma (z + \alpha )}}{{\Gamma (z)}}. $

(2)

命题0.1^[21]

$ \frac{1}{{\Gamma (z)}} = z{{\rm{e}}^{\gamma z}}\prod\limits_{i = 1}^\infty \{ (1 + \frac{z}{n}){{\rm{e}}^{-z/n}}\} . $

(3)

式(3) 中, $\gamma$代表欧拉常数, 其具体定义为

$ \gamma = \mathop {\lim }\limits_{n \to \infty } (\sum\limits_{k = 1}^n {\frac{1}{k}}-\ln n). $

命题0.2^[21] (欧拉反射公式)

$ \Gamma (z)\Gamma (1-z) = \frac{\pi }{{\sin \pi z}}. $

(4)

定义0.3 双伽马函数$\psi(z)$定义为

$ \psi (z) = \frac{{\Gamma '(z)}}{{\Gamma (z)}}. $

(5)

命题0.3^[21] 双伽马函数$\psi(z)$有如下性质.

$ \psi (z + n) = \frac{1}{z} + \frac{1}{{z + 1}} + \cdots + \frac{1}{{z + n-1}} + \psi (z), \quad n = 1, 2, 3, \cdots, $

(6)

$ \psi (\frac{p}{q}) =-\gamma-\frac{\pi }{2}\cot \frac{{\pi p}}{q}-\ln q + 2\sum\limits_{n = 1}^{\left\lfloor {q/2} \right\rfloor } {\cos } \frac{{2\pi np}}{q}\ln (2\sin \frac{{n\pi }}{q}), $

(7)

其中$0 < p < q$; $\sum'$表示当$q$是偶数时, 指标为$n=q/2$的项要除以2.这里, $\lfloor q/2\rfloor$表示不大于$q/2$的最大整数.

公式(7) 也可以表达为

$ \psi (\frac{p}{q}) =-\gamma-\frac{\pi }{2}\cot \frac{{\pi p}}{q}-\ln q + \sum\limits_{n = 1}^{q - 1} {\cos } \frac{{2\pi np}}{q}\ln (2\sin \frac{{n\pi }}{q}). $

(8)

命题0.4^[20] (Dougall $_5F_4$公式)若$\mathfrak{R}(a+b+c+d+1)>0$, 则

$ \begin{array}{l} \sum\limits_{n = 0}^\infty {\frac{{(a + 2n)\Gamma (a + n)\Gamma (n-b)\Gamma (n-c)\Gamma (n-d)}}{{n!\Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)}}} \\ = \frac{{\Gamma ( - b)\Gamma ( - c)\Gamma ( - d)\Gamma (a + b + c + d + 1)}}{{\Gamma (a + b + c + 1)\Gamma (a + b + d + 1)\Gamma (a + c + d + 1)}}. \end{array} $

(9)

命题0.5^[20] 若$\mathfrak{R}(a+b+c+d+\alpha-\beta-\gamma-\delta+1)>0$, 则

$ \begin{array}{l} \sum\limits_{n = 0}^\infty {\frac{{(\alpha + a + 2n){{(\alpha )}_{a + n}}{{(\beta )}_{n-b}}{{(\gamma )}_{n-c}}{{(\delta )}_{n-d}}}}{{n!{{(1 + \alpha - \beta )}_{a + b + n}}{{(1 + \alpha - \gamma )}_{a + c + n}}{{(1 + \alpha - \delta )}_{a + d + n}}}}} \\ = \frac{{\Gamma (1 + \alpha - \beta )\Gamma (1 + \alpha - \gamma )\Gamma (1 + \alpha - \delta )\Gamma (2 + \alpha - \beta - \gamma - \delta )}}{{\Gamma (\alpha )\Gamma (1 + \alpha - \beta - \gamma )\Gamma (1 + \alpha - \beta - \delta )\Gamma (1 + \alpha - \gamma - \delta )}}\\ \times \frac{{{{(\beta )}_{ - b}}{{(\gamma )}_{ - c}}{{(\delta )}_{ - d}}{{(2 + \alpha - \beta - \gamma - \delta )}_{a + b + c + d - 1}}}}{{{{(1 + \alpha - \beta - \gamma )}_{a + b + c}}{{(1 + \alpha - \beta - \delta )}_{a + b + d}}{{(1 + \alpha - \gamma - \delta )}_{a + c + d}}}}. \end{array} $

(10)

1 Dougall $_5F_4$求和公式的推广以及$1/\pi$的Ramanujan型级数

引理1.1 对于任意的复数$x, y$, 我们有

$ \mathop {\lim }\limits_{n \to \infty } \frac{{\Gamma (n + x)}}{{\Gamma (n + y)}}{n^{y-x}} = 1. $

证 明 直接利用式(2) 和定义0.1便可证明该引理.

定理1.1 若${\mathfrak{R}}(a+b+c+d+1)>0$, 则

$ \begin{array}{l} \frac{{{{\rm{d}}^k}}}{{{\rm{d}}{a^k}}}[\sum\limits_{n = 0}^\infty {\frac{{(a + 2n)\Gamma (a + n)\Gamma (n-b)\Gamma (n-c)\Gamma (n-d)}}{{n!\Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)}}}]\\ = \sum\limits_{n = 0}^\infty {\frac{{{{\rm{d}}^k}}}{{{\rm{d}}{a^k}}}} [\frac{{(a + 2n)\Gamma (a + n)\Gamma (n-b)\Gamma (n-c)\Gamma (n-d)}}{{n!\Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)}}]. \end{array} $

(11)

证 明 显然, 上式左边被求导的级数是绝对收敛的.接下来我们需要证明此级数也是一致收敛的.记

$ {u_n}(a, b, c, d) = \frac{{(a + 2n)\Gamma (a + n)\Gamma (n-b)\Gamma (n-c)\Gamma (n-d)}}{{n!\Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)}}. $

利用引理1.1, 我们有

$ \begin{array}{l} \mathop {\lim }\limits_{n \to \infty } \frac{{\Gamma (n + a)}}{{\Gamma (n + a + b + 1)}}{n^{b + 1}} = 1;\quad \mathop {\lim }\limits_{n \to \infty } \frac{{\Gamma (n-c)}}{{\Gamma (n + c + a + 1)}}{n^{a + 2c + 1}} = 1;\\ \mathop {\lim }\limits_{n \to \infty } \frac{{\Gamma (n-d)}}{{\Gamma (n + d + a + 1)}}{n^{a + 2d + 1}} = 1;\quad \mathop {\lim }\limits_{n \to \infty } \frac{{\Gamma (n-b)}}{{\Gamma (n + 1)}}{n^{b + 1}} = 1. \end{array} $

因此, 当$n\rightarrow\infty$时,

$ {u_n}(a,b,c,d) \sim \frac{{a + 2n}}{{n \cdot {n^b}}} \cdot \frac{1}{{{n^{b + 1}}}} \cdot \frac{1}{{{n^{a + 2c + 1}}}} \cdot \frac{1}{{{n^{a + 2d + 1}}}} \sim \frac{2}{{{n^{2a + 2b + 2c + 2d + 3}}}}. $

又由于${\mathfrak{R}}(2a+2b+2c+2d+3)>1$, 故此级数一致收敛.

定义1.1 设$a, b, c, d$为复数, $n$为整数.我们定义

$ \begin{array}{l} H(a, b, c, d, n): = [\frac{1}{{a + 2n}} + \psi (a + n)]\\ - [\psi (a + b + n + 1) + \psi (a + c + n + 1) + \psi (a + d + n + 1)], \end{array} $

其中$\psi$是由式(5) 定义的双伽马函数.

定理1.2 若${\mathfrak{R}}(a+b+c+d+1)>0$, 则

$ \begin{array}{l} \sum\limits_{n = 0}^\infty H (a, b, c, d, n)\frac{{(a + 2n)\Gamma (a + n)\Gamma (n- b)\Gamma (n- c)\Gamma (n- d)}}{{n!\Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)}}\\ = \frac{{\Gamma ( - b)\Gamma ( - c)\Gamma ( - d)\Gamma (a + b + c + d + 1)}}{{\Gamma (a + b + c + 1)\Gamma (a + b + d + 1)\Gamma (a + c + d + 1)}}\{ \psi (a + b + c + d + 1)\\ - [\psi (a + b + c + 1) + \psi (a + b + d + 1) + \psi (a + c + d + 1)]\} . \end{array} $

(12)

证 明 首先, 我们有

$ \frac{{\rm{d}}}{{{\rm{d}}a}}[\frac{{(a + 2n)\Gamma (a + n)}}{{\Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)}}] = \frac{{{N_1}(a)}}{{{D_1}(a)}}, $

其中

$ \begin{array}{l} {N_1}(a) = [\Gamma (a + n) + (a + 2n)\Gamma '(a + n)]\Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)\\ - (a + 2n)\Gamma (a + n)[\Gamma '(a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)\\ + \Gamma (a + b + n + 1)\Gamma '(a + c + n + 1)\Gamma (a + d + n + 1)\\ + \Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma '(a + d + n + 1)], \end{array} $

而

$ {D_1}(a) = {[\Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)]^2}. $

因此

$ \begin{array}{l} \frac{{\rm{d}}}{{{\rm{d}}a}}[\frac{{(a + 2n)\Gamma (a + n)\Gamma (n-b)\Gamma (n-c)\Gamma (n-d)}}{{n!\Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)}}]\\ = \frac{{[\frac{1}{{a + 2n}} + \psi (a + n)](a + 2n)\Gamma (a + n)\Gamma (n - b)\Gamma (n - c)\Gamma (n - d)}}{{n!\Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)}}\\ - \frac{{(a + 2n)\Gamma (a + n)\Gamma (n - b)\Gamma (n - c)\Gamma (n - d)}}{{n!\Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)}}\\ \times [\psi (a + b + n + 1) + \psi (a + c + n + 1) + \psi (a + d + n + 1)] = \\ H(a, b, c, d, n)\frac{{(a + 2n)\Gamma (a + n)\Gamma (n -b)\Gamma (n -c)\Gamma (n -d)}}{{n!\Gamma (a + b + n + 1)\Gamma (a + c + n + 1)\Gamma (a + d + n + 1)}}. \end{array} $

(13)

另一方面

$ \frac{{\rm{d}}}{{{\rm{d}}a}}[\frac{{\Gamma (-b)\Gamma (-c)\Gamma (-d)\Gamma (a + b + c + d + 1)}}{{\Gamma (a + b + c + 1)\Gamma (a + b + d + 1)\Gamma (a + c + d + 1)}}] = \frac{{{N_2}(a)}}{{{D_2}(a)}}, $

其中

$ \begin{array}{l} {N_2}(a) = \Gamma (- b)\Gamma (- c)\Gamma (- d)\Gamma '(a + b + c + d + 1)\\ \Gamma (a + b + c + 1)\Gamma (a + b + d + 1)\Gamma (a + c + d + 1)\\ - \Gamma ( - b)\Gamma ( - c)\Gamma ( - d)\Gamma (a + b + c + d + 1)\\ [\Gamma '(a + b + c + 1)\Gamma (a + b + d + 1)\Gamma (a + c + d + 1)\\ + \Gamma (a + b + c + 1)\Gamma '(a + b + d + 1)\Gamma (a + c + d + 1)\\ + \Gamma (a + b + c + 1)\Gamma (a + b + d + 1)\Gamma '(a + c + d + 1)], \end{array} $

而

$ {D_2}(a) = {[\Gamma (a + b + c + 1)\Gamma (a + b + d + 1)\Gamma (a + c + d + 1)]^2}. $

根据$N_2(a)$和$D_2(a)$的表达式, 可进一步得到

$ \begin{array}{l} \frac{{{N_2}(a)}}{{{D_2}(a)}} = \frac{{\Gamma (- b)\Gamma (- c)\Gamma (- d)\Gamma (a + b + c + d + 1)}}{{\Gamma (a + b + c + 1)\Gamma (a + b + d + 1)\Gamma (a + c + d + 1)}}\{ \psi (a + b + c + d + 1)\\ - [\psi (a + b + c + 1) + \psi (a + b + d + 1) + \psi (a + c + d + 1)]\} . \end{array} $

(14)

最终, 结合公式(9)、(11)、(13) 以及(14), 我们完成了公式(12) 的证明.

定理1.3 若${\mathfrak{R}}(a+b+c+d+\alpha-\beta-\gamma-\delta+1)>0$, 则

$ \begin{array}{l} \sum\limits_{n = 0}^\infty H (a + \alpha, b- \beta, c- \gamma, d- \delta, n)\frac{{(a + 2n + \alpha ){{(\alpha )}_{a + n}}{{(\beta )}_{n - b}}{{(\gamma )}_{n - c}}{{(\delta )}_{n - d}}}}{{n!{{(1 + \alpha - \beta )}_{a + b + n}}{{(1 + \alpha - \gamma )}_{a + c + n}}{{(1 + \alpha - \delta )}_{a + d + n}}}}\\ = \frac{{\Gamma (1 + \alpha - \beta )\Gamma (1 + \alpha - \gamma )\Gamma (1 + \alpha - \delta )\Gamma (2 + \alpha - \beta - \gamma - \delta )}}{{\Gamma (\alpha )\Gamma (1 + \alpha - \beta - \gamma )\Gamma (1 + \alpha - \beta - \delta )\Gamma (1 + \alpha - \gamma - \delta )}}\\ \times \frac{{{{(\beta )}_{ - b}}{{(\gamma )}_{ - c}}{{(\delta )}_{ - d}}{{(2 + \alpha - \beta - \gamma - \delta )}_{a + b + c + d - 1}}}}{{{{(1 + \alpha - \beta - \gamma )}_{a + b + c}}{{(1 + \alpha - \beta - \delta )}_{a + b + d}}{{(1 + \alpha - \gamma - \delta )}_{a + c + d}}}}\\ \times \{ \psi (a + b + c + d + 1 + \alpha - \beta - \gamma - \delta ) - [\psi (a + b + c + 1 + \alpha-\beta-\gamma )\\ + \psi (a + b + d + 1 + \alpha-\beta - \delta ) + \psi (a + c + d + 1 + \alpha - \gamma - \delta )]\} . \end{array} $

(15)

证 明 利用公式(2), 并将式(12) 中的$(a, b, c, d)$替换为$(a+\alpha, b-\beta, c-\gamma, d-\delta)$, 就完成了定理的证明.

定理1.4 若${\mathfrak{R}}(a+b+c+d)>0$, 则

$ \begin{array}{l} \sum\limits_{n = 0}^\infty H (a + \frac{1}{2}, b- \frac{1}{2}, c- \frac{1}{3}, d- \frac{2}{3}, n)\frac{{(4n + 2a + 1){{(\frac{1}{2})}_{n + a}}{{(\frac{1}{2})}_{n - b}}{{(\frac{1}{3})}_{n - c}}{{(\frac{2}{3})}_{n - d}}}}{{n!{{(1)}_{a + b + n}}{{(\frac{7}{6})}_{a + c + n}}{{(\frac{5}{6})}_{a + d + n}}}}\\ = \frac{{{{(\frac{1}{2})}_{ - b}}{{(\frac{1}{3})}_{ - c}}{{(\frac{2}{3})}_{ - d}}{{(1)}_{a + b + c + d - 1}}}}{{\pi \sqrt 3 {{(\frac{2}{3})}_{a + b + c}}{{(\frac{1}{3})}_{a + b + d}}{{(\frac{1}{2})}_{a + c + d}}}}\{ \psi (a + b + c + d) - [\psi (a + b + c + \frac{2}{3}){\mkern 1mu} \\ + \psi (a + b + d + \frac{1}{3}) + \psi (a + c + d + \frac{1}{2})]\} . \end{array} $

(16)

证 明 在式(15) 中令$(\alpha, \beta, \gamma, \delta)=(\frac{1}{2}, \frac{1}{2}, \frac{1}{3}, \frac{2}{3})$可得

$ \begin{array}{l} \sum\limits_{n = 0}^\infty H (a + \frac{1}{2}, b- \frac{1}{2}, c- \frac{1}{3}, d- \frac{2}{3}, n)\frac{{(4n + 2a + 1){{(\frac{1}{2})}_{n + a}}{{(\frac{1}{2})}_{n - b}}{{(\frac{1}{3})}_{n - c}}{{(\frac{2}{3})}_{n - d}}}}{{n!{{(1)}_{a + b + n}}{{(\frac{7}{6})}_{a + c + n}}{{(\frac{5}{6})}_{a + d + n}}}}\\ = \frac{{\Gamma (\frac{7}{6})\Gamma (\frac{5}{6}){{(\frac{1}{2})}_{ - b}}{{(\frac{1}{3})}_{ - c}}{{(\frac{2}{3})}_{ - d}}{{(1)}_{a + b + c + d - 1}}}}{{{\Gamma ^2}(\frac{1}{2})\Gamma (\frac{2}{3})\Gamma (\frac{1}{3}){{(\frac{2}{3})}_{a + b + c}}{{(\frac{1}{3})}_{a + b + d}}{{(\frac{1}{2})}_{a + c + d}}}}\{ \psi (a + b + c + d)\\ - [\psi (a + b + c + \frac{2}{3}) + \psi (a + b + d + \frac{1}{3}) + \psi (a + c + d + \frac{1}{2})]\} . \end{array} $

(17)

由于

$ \Gamma (\frac{7}{6}) = \Gamma (1 + \frac{1}{6}) = \frac{1}{6}\Gamma (\frac{1}{6}), $

(18)

利用欧拉反射公式(4), 我们有

$ \Gamma (\frac{1}{6})\Gamma (\frac{5}{6}) = \frac{\pi }{{\sin (\frac{\pi }{6})}} = 2\pi, $

(19)

以及

$ \Gamma (\frac{1}{3})\Gamma (\frac{2}{3}) = \frac{\pi }{{\sin (\frac{\pi }{3})}} = \frac{{2\pi }}{{\sqrt 3 }}. $

(20)

在等式(17) 的两边同时乘以$2$且将式(18)-(20) 代入到此公式, 我们就得到了公式(16).

例1 在定理1.4中, 取$(a, b, c, d)=(1, 0, 0, 0)$, 则有

$ \begin{array}{l} \sum\limits_{n = 0}^\infty H (\frac{3}{2}, - \frac{1}{2}, - \frac{1}{3}, - \frac{2}{3}, n)\frac{{(4n + 3){{(\frac{1}{2})}_{n + 1}}{{(\frac{1}{2})}_n}{{(\frac{1}{3})}_n}{{(\frac{2}{3})}_n}}}{{n!{{(1)}_{n + 1}}{{(\frac{7}{6})}_{n + 1}}{{(\frac{5}{6})}_{n + 1}}}}\\ = \frac{{\psi (1) - [\psi (\frac{5}{3}) + \psi (\frac{4}{3}) + \psi (\frac{3}{2})]}}{{\pi \sqrt 3 (\frac{2}{3})(\frac{1}{3})(\frac{1}{2})}}. \end{array} $

利用式(6) 和(8) 来计算$\psi(\frac{5}{3})$, $\psi(\frac{4}{3})$, $\psi(\frac{3}{2})$的值, 则能够进一步得到

$ \begin{array}{l} \frac{{3\sqrt 3 }}{\pi }[2\gamma-\frac{{13}}{2} + 3\ln 3 + 2\ln 2]\\ = \sum\limits_{n = 0}^\infty \{ [\frac{2}{{4n + 3}} + \psi (n + \frac{3}{2})] - [\psi (n + 2) + \psi (n + \frac{{13}}{6}) + \psi (n + \frac{{11}}{6})]\} \\ \times \frac{{(4n + 3){{(\frac{1}{2})}_{n + 1}}{{(\frac{1}{2})}_n}{{(\frac{1}{3})}_n}{{(\frac{2}{3})}_n}}}{{n!{{(1)}_{n + 1}}{{(\frac{7}{6})}_{n + 1}}{{(\frac{5}{6})}_{n + 1}}}}. \end{array} $

例2 在定理1.4中令$(a, b, c, d)=(1, -1, 1, 0)$, 可以得出

$ \begin{array}{l} \frac{{\sqrt 3 }}{{2\pi }}[\frac{{25}}{6}-2(\ln 2 + \gamma )-3\ln 3]\\ = \sum\limits_{n = 0}^\infty \{ [\frac{2}{{4n + 3}} + {\psi _0}(n + \frac{3}{2})] - [\psi (n + 1) + \psi (n + \frac{{19}}{6}) + \psi (n + \frac{{11}}{6})]\} {\mkern 1mu} \\ \times \frac{{(4n + 3)(\frac{1}{2})_{n + 1}^2{{(\frac{1}{3})}_{n -1}}{{(\frac{2}{3})}_n}}}{{n{!^2}{{(\frac{7}{6})}_{n + 2}}{{(\frac{5}{6})}_{n + 1}}}}. \end{array} $

2 $\pi^2$的一般级数展开式

在这一节中, 我们利用定理1.3来证明下面的级数展开式.

定理2.1 若${\mathfrak{R}}(a+b+c+d-\frac{1}{2})>0$, 则

$ \begin{array}{l} \frac{{{\pi ^2}{{(\frac{1}{2})}_{- b}}{{(\frac{1}{2})}_{- c}}{{(\frac{1}{2})}_{- d}}{{(\frac{1}{2})}_{a + b + c + d - 1}}}}{{{{(1)}_{a + b + c - 1}}{{(1)}_{a + b + d - 1}}{{(1)}_{a + c + d - 1}}}}\{ \psi (a + b + c + d - \frac{1}{2})\\ - [\psi (a + b + c) + \psi (a + b + d) + \psi (a + c + d)]\} \\ = \sum\limits_{n = 0}^\infty H (a, b -\frac{1}{2}, c -\frac{1}{2}, d -\frac{1}{2}, n)\frac{{(a + 2n){{(1)}_{a + n - 1}}{{(\frac{1}{2})}_{n - b}}{{(\frac{1}{2})}_{n - c}}{{(\frac{1}{2})}_{n - d}}}}{{n!{{(\frac{1}{2})}_{a + b + n}}{{(\frac{1}{2})}_{a + c + n}}{{(\frac{1}{2})}_{a + d + n}}}}. \end{array} $

证 明 利用递归关系式$\Gamma(z+1)=z\Gamma(z)$, 我们可将定理1.3中的(15) 式表示为

$ \begin{array}{l} \sum\limits_{n = 0}^\infty H (a + \alpha, b- \beta, c- \gamma, d- \delta, n)\frac{{(a + 2n + \alpha ){{(\alpha + 1)}_{a + n - 1}}{{(\beta )}_{n - b}}{{(\gamma )}_{n - c}}{{(\delta )}_{n - d}}}}{{n!{{(1 + \alpha - \beta )}_{a + b + n}}{{(1 + \alpha - \gamma )}_{a + c + n}}{{(1 + \alpha - \delta )}_{a + d + n}}}}\\ = \frac{{\Gamma (1 + \alpha - \beta )\Gamma (1 + \alpha - \gamma )\Gamma (1 + \alpha - \delta )\Gamma (2 + \alpha - \beta - \gamma - \delta )}}{{\Gamma (\alpha + 1)\Gamma (2 + \alpha - \beta - \gamma )\Gamma (2 + \alpha - \beta - \delta )\Gamma (2 + \alpha - \gamma - \delta )}}\\ \times \frac{{{{(\beta )}_{ - b}}{{(\gamma )}_{ - c}}{{(\delta )}_{ - d}}{{(2 + \alpha - \beta - \gamma - \delta )}_{a + b + c + d - 1}}}}{{{{(2 + \alpha - \beta - \gamma )}_{a + b + c - 1}}{{(2 + \alpha - \beta - \delta )}_{a + b + d - 1}}{{(2 + \alpha - \gamma - \delta )}_{a + c + d - 1}}}}\\ \times \{ \psi (a + b + c + d + 1 + \alpha - \beta - \gamma - \delta ) - [\psi (a + b + c + 1 + \alpha-\beta-\gamma )\\ + \psi (a + b + d + 1 + \alpha-\beta - \delta ) + \psi (a + c + d + 1 + \alpha - \gamma - \delta )]\} . \end{array} $

在上面等式中令$(\alpha, \beta, \gamma, \delta)=(0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$并将$\Gamma(\frac{1}{2})=\sqrt{\pi}$代入所得到的结果, 我们便完成了定理的证明.

例3 在定理2.1中, 令$(a, b, c, d)=(1, 0, 0, 0)$, 则有

$ \frac{{{\pi ^2}}}{4}[\gamma-\ln 2] = \sum\limits_{n = 0}^\infty [\frac{1}{{2n + 1}} + \psi (n + 1)-3\psi (n + \frac{3}{2})]\frac{1}{{{{(2n + 1)}^2}}}. $

例4 在定理2.1中, 取$(a, b, c, d)=(1, 1, 0, 0)$, 则有

$ \begin{array}{l} \frac{{{\pi ^2}}}{{16}}[\ln 2-\gamma] = \sum\limits_{n = 0}^\infty [\frac{1}{{2n + 1}} + \psi (n + 1)-\psi (n + \frac{5}{2})-2\psi (n + \frac{3}{2})]\\ \times \frac{1}{{(2n -1){{(2n + 1)}^2}(2n + 3)}}. \end{array} $

例5 在定理2.1中, 令$(a, b, c, d)=(2, -1, 0, 0)$, 则可得

$ \begin{array}{l} \frac{{{\pi ^2}}}{{64}}[2(\gamma-\ln 2)-1] = \sum\limits_{n = 0}^\infty [\frac{1}{{2(n + 1)}} + \psi (n + 2)-\psi (n + \frac{3}{2})-2\psi (n + \frac{5}{2})]\\ \times \frac{{{{(n + 1)}^2}}}{{{{(2n + 1)}^2}{{(2n + 3)}^2}}}. \end{array} $

3 $\Gamma^{-3}(\frac{2}{3})$的一般级数展开式

定理3.1 若${\mathfrak{R}}(a+b+c+d+\frac{1}{3})>0$, 则有

$ \begin{array}{l} \frac{{3{{(\frac{1}{3})}_{- b}}{{(\frac{1}{3})}_{- c}}{{(\frac{1}{3})}_{- d}}{{(\frac{1}{3})}_{a + b + c + d}}}}{{{\Gamma ^3}(\frac{2}{3}){{(\frac{2}{3})}_{a + b + c}}{{(\frac{2}{3})}_{a + b + d}}{{(\frac{2}{3})}_{a + c + d}}}}\\ \times \{ \psi (a + b + c + d + \frac{1}{3}) - [\psi (a + b + c + \frac{2}{3}) + \psi (a + b + d + \frac{2}{3}) + \psi (a + c + d + \frac{2}{3})]\} \\ = \sum\limits_{n = 0}^\infty H (a + \frac{1}{3}, b -\frac{1}{3}, c -\frac{1}{3}, d -\frac{1}{3}, n)\frac{{(3a + 6n + 1){{(\frac{1}{3})}_{a + n}}{{(\frac{1}{3})}_{n - b}}{{(\frac{1}{3})}_{n - c}}{{(\frac{1}{3})}_{n - d}}}}{{n!{{(1)}_{a + b + n}}{{(1)}_{a + c + n}}{{(1)}_{a + d + n}}}}. \end{array} $

(21)

证 明 在定理1.3中令$(\alpha, \beta, \gamma, \delta)=(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, \frac{1}{3})$, 并利用关系$\Gamma(\frac{4}{3})=\frac{1}{3}\Gamma(\frac{1}{3})$以及$(\frac{1}{3})(\frac{4}{3})_{a+b+c+d-1}=(\frac{1}{3})_{a+b+c+d}$, 我们便完成了定理的证明.

例6 在定理3.1中令$(a, b, c, d)=(0, 0, 0, 0)$, 我们可推出

$ \frac{3}{{{\Gamma ^3}(\frac{2}{3})}}[2\gamma-\frac{{2\pi \sqrt 3 }}{3} + 3\ln 3] = \sum\limits_{n = 0}^\infty [\frac{3}{{6n + 1}} + \psi (n + \frac{1}{3})-3\psi (n + 1)]\frac{{(6n + 1)(\frac{1}{3})_n^4}}{{n{!^4}}}. $

例7 在定理3.1中令$(a, b, c, d)=(1, 0, 0, 0)$, 我们可得到结论

$ \begin{array}{l} \frac{{27}}{{16{\Gamma ^3}(\frac{2}{3})}}[2\gamma + 3\ln 3-\frac{{2\pi \sqrt 3 }}{3}-\frac{3}{2}]\\ = \sum\limits_{n = 0}^\infty [\frac{3}{{6n + 4}} + \psi (n + \frac{4}{3})-3\psi (n + 2)]\frac{{(3n + 2){{(\frac{1}{3})}_{n + 1}}(\frac{1}{3})_n^3}}{{n!(n + 1){!^3}}}. \end{array} $

例8 在定理3.1中选取$(a, b, c, d)=(1, -1, 0, 0)$, 我们可得到

$ \begin{array}{l} \frac{3}{{4{\Gamma ^3}(\frac{2}{3})}}[2\gamma + 3\ln 3-\frac{{2\pi \sqrt 3 }}{3}-\frac{3}{2}]\\ = \sum\limits_{n = 0}^\infty [\frac{3}{{6n + 4}} + \psi (n + \frac{4}{3})-\psi (n + 1)-2\psi (n + 2)]\frac{{(3n + 2)(\frac{1}{3})_{n + 1}^2(\frac{1}{3})_n^2}}{{n{!^2}(n + 1){!^2}}}. \end{array} $

4 $\frac{1}{\sqrt{\pi}}\Gamma^{-2}(\frac{3}{4})$的一般级数展开式

定理4.1 若${\mathfrak{R}}(a+b+c+d+\frac{1}{2})>0$, 则有

$ \begin{array}{l} \frac{{2\sqrt 2 {{(\frac{1}{4})}_{- b}}{{(\frac{1}{4})}_{- c}}{{(\frac{1}{4})}_{- d}}{{(\frac{1}{2})}_{a + b + c + d}}}}{{\sqrt \pi {\Gamma ^2}(\frac{3}{4}){{(\frac{3}{4})}_{a + b + c}}{{(\frac{3}{4})}_{a + b + d}}{{(\frac{3}{4})}_{a + c + d}}}}\\ \times \{ \psi (a + b + c + d + \frac{1}{2}) - [\psi (a + b + c + \frac{3}{4}) + \psi (a + b + d + \frac{3}{4}) + \psi (a + c + d + \frac{3}{4})]\} \\ = \sum\limits_{n = 0}^\infty H (a + \frac{1}{4}, b -\frac{1}{4}, c -\frac{1}{4}, d -\frac{1}{4}, n)\frac{{(4a + 8n + 1){{(\frac{1}{4})}_{a + n}}{{(\frac{1}{4})}_{n - b}}{{(\frac{1}{4})}_{n - c}}{{(\frac{1}{4})}_{n - d}}}}{{n!{{(1)}_{a + b + n}}{{(1)}_{a + c + n}}{{(1)}_{a + d + n}}}}. \end{array} $

(22)

证 明 在定理1.3中令$(\alpha, \beta, \gamma, \delta)=(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4})$, 并将$\Gamma(\frac{1}{2})=\sqrt{\pi}$和$\Gamma(\frac{1}{4})\Gamma (\frac{3}{4})=\sqrt{2}\pi$代入到所得的等式中, 我们就完成了定理的证明.

例9 在定理4.1中令$a=b=c=d=0$, 则有

$ \begin{array}{l} \frac{{2\sqrt 2 }}{{\sqrt \pi {\Gamma ^2}(\frac{3}{4})}}[2\gamma + 7\ln 2-\frac{{3\pi }}{2}]\\ = \sum\limits_{n = 0}^\infty [\frac{4}{{8n + 1}} + \psi (n + \frac{1}{4})-3\psi (n + 1)]\frac{{(8n + 1)(\frac{1}{4})_n^4}}{{n{!^4}}}. \end{array} $

例10 定理4.1中令$a=1$且$b=c=d=0$, 则有

$ \begin{array}{l} \frac{{64\sqrt 2 }}{{27\sqrt \pi {\Gamma ^2}(\frac{3}{4})}}[2\gamma + 7\ln 2-2-\frac{{3\pi }}{2}]\\ = \sum\limits_{n = 0}^\infty [\frac{4}{{8n + 5}} + \psi (n + \frac{5}{4})-3\psi (n + 2)]\frac{{(8n + 5){{(\frac{1}{4})}_{n + 1}}(\frac{1}{4})_n^3}}{{n!(n + 1){!^3}}}. \end{array} $

5 $1/\pi^2$的Ramanujan型级数

定理5.1 若${\mathfrak{R}}(a+b+c+d)>0$, 则有

$ \begin{array}{l} \frac{{2{{(\frac{1}{2})}_{- b}}{{(\frac{1}{2})}_{- c}}{{(\frac{1}{2})}_{- d}}{{(1)}_{a + b + c + d - 1}}}}{{{\pi ^2}{{(\frac{1}{2})}_{a + b + c}}{{(\frac{1}{2})}_{a + b + d}}{{(\frac{1}{2})}_{a + c + d}}}}{\mkern 1mu} \\ \times \{ \psi (a + b + c + d) - [\psi (a + b + c + \frac{1}{2}) + \psi (a + b + d + \frac{1}{2}) + \psi (a + c + d + \frac{1}{2})]\} \\ = \sum\limits_{n = 0}^\infty H (a + \frac{1}{2}, b -\frac{1}{2}, c -\frac{1}{2}, d -\frac{1}{2}, n)\frac{{(4n + 2a + 1){{(\frac{1}{2})}_{a + n}}{{(\frac{1}{2})}_{n - b}}{{(\frac{1}{2})}_{n - c}}{{(\frac{1}{2})}_{n - d}}}}{{n!{{(1)}_{a + b + n}}{{(1)}_{a + c + n}}{{(1)}_{a + d + n}}}}. \end{array} $

(23)

证 明 在定理1.3中取$(\alpha, \beta, \gamma, \delta)=(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})$并将$\Gamma(\frac{1}{2})=\sqrt{\pi}$应用到所得到的等式中, 我们就完成了定理的证明.

若在定理5.1中令$b=c=d=-a$, 则可得到以下推论.

推论5.1 若${\mathfrak{R}}(a) < 0$, 则

$ \begin{array}{l} \frac{{2(\frac{1}{2})_a^3{{(1)}_{- 2a- 1}}}}{{{\pi ^2}(\frac{1}{2})_{- a}^3}}[\psi (-2a)-3\psi (-a + \frac{1}{2})]\\ = \sum\limits_{n = 0}^\infty H (a + \frac{1}{2}, -a -\frac{1}{2}, -a - \frac{1}{2}, - a - \frac{1}{2}, n)\frac{{(4n + 2a + 1)(\frac{1}{2})_{a + n}^4}}{{n{!^4}}}. \end{array} $

(24)

例11 在推论5.1中令$a=-1$, 则

$ \frac{{128}}{{{\pi ^2}}}[5-2\gamma-6\ln 2] = \sum\limits_{n = 0}^\infty [\frac{2}{{4n-1}} + \psi (n-\frac{1}{2})-3\psi (n + 1)]\frac{{(4n -1)(\frac{1}{2})_{n -1}^4}}{{n{!^4}}}. $

例12 在推论5.1中令$a=-2$, 则

$ \frac{{16384}}{{243{\pi ^2}}}[2\gamma + 6\ln 2-\frac{{37}}{6}] = \sum\limits_{n = 0}^\infty [\frac{2}{{4n-3}} + \psi (n-\frac{3}{2})-3\psi (n + 1)]\frac{{(4n -3)(\frac{1}{2})_{n -2}^4}}{{n{!^4}}}. $

下面, 我们将继续讨论定理5.1的特殊情形.在定理5.1中, 若我们选取$(a, b, c, d)=(k, 0, 0, 0)$并将其简化, 我们很容易得到下面的命题.

命题5.1 若$k$是正整数, 则我们有

$ \begin{array}{l} \frac{{2(k- 1)!}}{{{\pi ^2}(\frac{1}{2})_k^3}}[\psi (k)-3\psi (k + \frac{1}{2})]\\ = \sum\limits_{n = 0}^\infty H (k + \frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, n)\frac{{(4n + 2k + 1){{(\frac{1}{2})}_{n + k}}(\frac{1}{2})_n^3}}{{n!(n + k){!^3}}}. \end{array} $

(25)

例13 当$k=1$时, 我们可以得到

$ \frac{{64}}{{{\pi ^2}}}[\gamma + 3\ln 2-3] = \sum\limits_{n = 0}^\infty [\frac{2}{{4n + 3}} + \psi (n + \frac{3}{2})-3\psi (n + 2)]\frac{{(2n + 1)(4n + 3)(\frac{1}{2})_n^4}}{{{{(n + 1)}^3}n{!^4}}}. $

命题5.2 若$k$是一个非负整数, 则我们有以下公式

$ \begin{array}{l} \frac{{4k!}}{{{\pi ^2}{{(\frac{1}{2})}_k}(\frac{1}{2})_{k + 1}^2}}[2\psi (k + \frac{3}{2}) + \psi (k + \frac{1}{2})-\psi (k + 1)]\\ = \sum\limits_{n = 0}^\infty H (k + \frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, n)\frac{{(4n + 2k + 1){{(\frac{1}{2})}_{n + k}}{{(\frac{1}{2})}_{n -1}}(\frac{1}{2})_n^2}}{{n!{{(1)}_{n + k + 1}}(1)_{n + k}^2}}. \end{array} $

(26)

证 明 在定理5.1中令$(a, b, c, d)=(k, 1, 0, 0)$即可得到上面的等式.

例14 在命题5.2中令$k=0$, 我们可得到

$ \begin{array}{l} \frac{{32}}{{{\pi ^2}}}[2-\gamma-3\ln 2] = \sum\limits_{n = 0}^\infty [\frac{2}{{4n + 1}} + \psi (n + \frac{1}{2})-\psi (n + 2)-2\psi (n + 1)]\\ \times \frac{{(4n + 1){{(\frac{1}{2})}_{n -1}}(\frac{1}{2})_n^3}}{{(n + 1)n{!^4}}}. \end{array} $

例15 在命题5.2中取$k=1$, 我们可推出

$ \begin{array}{l} \frac{{128}}{{9{\pi ^2}}}[\frac{{19}}{3}-2\gamma-6\ln 2] = \sum\limits_{n = 0}^\infty [\frac{2}{{4n + 3}} + \psi (n + \frac{3}{2})-2\psi (n + 2)-\psi (n + 3)]\\ \times \frac{{(4n + 3){{(\frac{1}{2})}_{n + 1}}{{(\frac{1}{2})}_{n -1}}(\frac{1}{2})_n^2}}{{n!(n + 2)!(n + 1){!^2}}}. \end{array} $

在定理5.1中选取$(a, b, c, d)=(k, 1, 1, 0)$, 我们有以下等式

$ \begin{array}{l} \frac{{2(\frac{1}{2})_{- 1}^2{{(1)}_{k + 1}}}}{{{\pi ^2}{{(\frac{1}{2})}_{k + 2}}(\frac{1}{2})_{k + 1}^2}}[\psi (k + 2)-\psi (k + \frac{5}{2})-2\psi (k + \frac{3}{2})]\\ = \sum\limits_{n = 0}^\infty H (k + \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, n)\frac{{(4n + 2k + 1){{(\frac{1}{2})}_{n + k}}(\frac{1}{2})_{n -1}^2{{(\frac{1}{2})}_n}}}{{n!(1)_{n + k + 1}^2{{(1)}_{n + k}}}}. \end{array} $

因此我们可得到下面的命题.

命题5.3 $k\geq0$是整数, 则我们有以下公式

$ \begin{array}{l} \frac{{8(k + 1)!}}{{{\pi ^2}{{(\frac{1}{2})}_{k + 2}}(\frac{1}{2})_{k + 1}^2}}[\psi (k + 2)-\psi (k + \frac{5}{2})-2\psi (k + \frac{3}{2})]\\ = \sum\limits_{n = 0}^\infty H (k + \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, n)\frac{{(4n + 2k + 1){{(\frac{1}{2})}_{n + k}}(\frac{1}{2})_{n -1}^2{{(\frac{1}{2})}_n}}}{{n!(1)_{n + k + 1}^2{{(1)}_{n + k}}}}. \end{array} $

(27)

例16 在式(27) 中令$k=0$, 我们可以得到

$ \begin{array}{l} \frac{{128}}{{3{\pi ^2}}}[2\gamma + 6\ln 2-\frac{{17}}{3}] = \sum\limits_{n = 0}^\infty [\frac{2}{{4n + 1}} + \psi (n + \frac{1}{2})-2\psi (n + 2)-\psi (n + 1)]\\ \times \frac{{(4n + 1)(\frac{1}{2})_{n -1}^2(\frac{1}{2})_n^2}}{{n{!^2}(n + 1){!^2}}}. \end{array} $

命题5.4 若$k$是整数, 且$k\geq-1$, 我们有

$ \begin{array}{l} \frac{{16(k + 2)!}}{{{\pi ^2}(\frac{1}{2})_{k + 2}^3}}[3\psi (k + \frac{5}{2})-\psi (k + 3)] = \sum\limits_{n = 0}^\infty H (k + \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, n)\\ \times \frac{{(4n + 2k + 1){{(\frac{1}{2})}_{n + k}}(\frac{1}{2})_{n -1}^3}}{{n!(n + k + 1){!^3}}}. \end{array} $

(28)

证 明 在定理5.1中取$(a, b, c, d)=(k, 1, 1, 1)$即完成证明.

例17 在式(28) 中令$k=-1$, 我们得到

$ \frac{{128}}{{{\pi ^2}}}[5-2\gamma-6\ln 2] = \sum\limits_{n = 0}^\infty [\frac{2}{{4n-1}} + \psi (n-\frac{1}{2})-3\psi (n + 1)]\frac{{(4n -1)(\frac{1}{2})_{n -1}^4}}{{n{!^4}}}. $

命题5.5 若$k$是整数, 且$k\geq2$, 我们有

$ \frac{{(k-2)!}}{{{\pi ^2}(\frac{1}{2})_{k-1}^2{{(\frac{1}{2})}_k}}} = \sum\limits_{n = 0}^\infty {\frac{{(4n + 2k + 1){{(\frac{1}{2})}_{n + k}}{{(\frac{1}{2})}_{n + 1}}(\frac{1}{2})_n^2}}{{n!(n + k-1)!(n + k){!^2}}}} . $

(29)

证 明 在文[20]的定理7.1中取$(a, b, c, d)=(k, -1, 0, 0)$.

例18 在式(29) 中令$k=2$, 可得

$ \frac{{16}}{{3{\pi ^2}}} = \sum\limits_{n = 0}^\infty {\frac{{(4n + 5)(\frac{1}{2})_n^2{{(\frac{1}{2})}_{n + 1}}{{(\frac{1}{2})}_{n + 2}}}}{{n!(n + 1)!(n + 2){!^2}}}} . $

例19 在式(29) 中令$k=3$, 可得

$ \frac{{128}}{{135{\pi ^2}}} = \sum\limits_{n = 0}^\infty {\frac{{(4n + 7)(\frac{1}{2})_n^2{{(\frac{1}{2})}_{n + 1}}{{(\frac{1}{2})}_{n + 3}}}}{{n!(n + 2)!(n + 3){!^2}}}} . $

命题5.6 若$k$是整数, 且$k\geq3$, 我们有

$ \frac{{(k-3)!}}{{2{\pi ^2}(\frac{1}{2})_{k-1}^2{{(\frac{1}{2})}_{k-2}}}} = \sum\limits_{n = 0}^\infty {\frac{{(4n + 2k + 1){{(\frac{1}{2})}_{n + k}}{{(\frac{1}{2})}_n}(\frac{1}{2})_{n + 1}^2}}{{n!(n + k - 1){!^2}(n + k)!}}} . $

(30)

证 明 在文[20]的定理7.1中令$(a, b, c, d)=(k, -1, -1, 0)$.

例20 在式(30) 中令$k=3$, 可得

$ \frac{{16}}{{9{\pi ^2}}} = \sum\limits_{n = 0}^\infty {\frac{{(4n + 7)(\frac{1}{2})_{n + 1}^2{{(\frac{1}{2})}_{n + 3}}{{(\frac{1}{2})}_n}}}{{n!(n + 3)!(n + 2){!^2}}}} . $

命题5.7 若$k$是整数, 且$k\geq-1$, 则

$ \frac{{16(k + 2)!}}{{{\pi ^2}(\frac{1}{2})_{k + 2}^3}} = \frac{{(16k + 8){{(\frac{1}{2})}_k}}}{{(k + 1){!^3}}}-\sum\limits_{n = 1}^\infty {\frac{{(4n + 2k + 1){{(\frac{1}{2})}_{n + k}}(\frac{1}{2})_{n-1}^3}}{{n!(n + k + 1){!^3}}}} . $

(31)

证 明 在文[20]的定理7.1中令$(a, b, c, d)=(k, 1, 1, 1)$.

例21 在式(31) 中取$k=-1$, 可得

$ \frac{{128}}{{{\pi ^2}}} = 16-\sum\limits_{n = 1}^\infty {\frac{{(4n-1)(\frac{1}{2})_{n-1}^4}}{{n{!^4}}}} . $

命题5.8 若$k$是整数, 且$k\geq2$, 则有

$ \begin{array}{l} \frac{{(k- 2)!}}{{{\pi ^2}(\frac{1}{2})_{k- 1}^2{{(\frac{1}{2})}_k}}}[\psi (k-1)-2\psi (k-\frac{1}{2}) - \psi (k + \frac{1}{2})]\\ = \sum\limits_{n = 0}^\infty H (k + \frac{1}{2}, -\frac{3}{2}, -\frac{1}{2}, -\frac{1}{2}, n)\frac{{(4n + 2k + 1){{(\frac{1}{2})}_{n + k}}{{(\frac{1}{2})}_{n + 1}}(\frac{1}{2})_n^2}}{{n!(n + k - 1)!(n + k){!^2}}}. \end{array} $

(32)

证 明 在定理5.1中令$(a, b, c, d)=(k, -1, 0, 0)$即完成证明.

例22 在式(32) 中取$k=2$, 则有

$ \begin{array}{l} \frac{{32}}{{3{\pi ^2}}}[\gamma + 3\ln 2-\frac{{10}}{3}] = \sum\limits_{n = 0}^\infty [\frac{2}{{4n + 5}} + \psi (n + \frac{5}{2})-\psi (n + 2)-2\psi (n + 3)]\\ \times \frac{{(4n + 5){{(\frac{1}{2})}_{n + 2}}{{(\frac{1}{2})}_{n + 1}}(\frac{1}{2})_n^2}}{{n!(n + 1)!(n + 2){!^2}}}. \end{array} $

命题5.9 若$k$是整数, 且$k\geq3$, 则

$ \begin{array}{l} \frac{{(k- 3)!}}{{2{\pi ^2}(\frac{1}{2})_{k- 1}^2{{(\frac{1}{2})}_{k- 2}}}}[\psi (k-2)-2\psi (k-\frac{1}{2}) - \psi (k - \frac{3}{2})]\\ = \sum\limits_{n = 0}^\infty H (k + \frac{1}{2}, -\frac{3}{2}, -\frac{3}{2}, -\frac{1}{2}, n)\frac{{(4n + 2k + 1){{(\frac{1}{2})}_{n + k}}(\frac{1}{2})_{n + 1}^2{{(\frac{1}{2})}_n}}}{{n!(n + k - 1){!^2}(n + k)!}}. \end{array} $

(33)

证 明 在定理5.1中令$(a, b, c, d)=(k, -1, -1, 0)$即完成证明.

例23 在式(33) 中取$k=3$便可推出下列等式

$ \begin{array}{l} \frac{{32}}{{9{\pi ^2}}}[\gamma + 3\ln 2-\frac{{11}}{3}] = \sum\limits_{n = 0}^\infty [\frac{2}{{4n + 7}} + \psi (n + \frac{7}{2})-\psi (n + 4)-2\psi (n + 3)]\\ \times \frac{{(4n + 7){{(\frac{1}{2})}_{n + 3}}(\frac{1}{2})_{n + 1}^2{{(\frac{1}{2})}_n}}}{{n!(n + 2){!^2}(n + 3)!}}. \end{array} $

命题5.10 若$k$是整数, 且$k\geq4$, 则有

$ \begin{array}{l} \frac{{(k- 4)!}}{{4{\pi ^2}(\frac{1}{2})_{k- 2}^3}}[\psi (k-3)-3\psi (k-\frac{3}{2})]\\ = \sum\limits_{n = 0}^\infty H (k + \frac{1}{2}, -\frac{3}{2}, -\frac{3}{2}, -\frac{3}{2}, n)\frac{{(4n + 2k + 1){{(\frac{1}{2})}_{n + k}}(\frac{1}{2})_{n + 1}^3}}{{n!(n + k - 1){!^3}}}. \end{array} $

(34)

证 明 在定理5.1中取$(a, b, c, d)=(k, -1, -1, -1)$即完成证明.

例24 在式(34) 中令$k=4$, 可推出以下等式

$ \frac{{32}}{{27{\pi ^2}}}[\gamma + 3\ln 2-4] = \sum\limits_{n = 0}^\infty [\frac{2}{{4n + 9}} + \psi (n + \frac{9}{2})-3\psi (n + 4)]\frac{{(4n + 9){{(\frac{1}{2})}_{n + 4}}(\frac{1}{2})_{n + 1}^3}}{{n!(n + 3){!^3}}}. $

致谢: 这是作者在华东师范大学攻读博士学位期间所写的论文.非常感谢导师刘治国教授在此期间给予的无私帮助和细心指导.