中心二项式系数在数论、统计学和概率论等数学领域有着重要应用. 1979年Apéry^[2]证明了下列无穷级数公式:

$ \sum\limits_{n = 1}^{\infty}\frac{(-1)^{n-1}}{n^3\big({}^{2n}_{\;n}\big)} = \frac{2}{5}\zeta(3), $

这里的$ \zeta(s) $为Riemann zeta函数,它的定义为

$ \zeta(s) = \sum\limits_{n = 1}^{\infty}\frac{1}{n^s}, \, \, \, \, \, \, \, \, \, {\cal R}(s)>1. $

Hurwitz zeta函数为Riemann zeta函数的一个推广:

$ \zeta(s, q) = \sum\limits_{k = 0}^{\infty}\frac{1}{(k+q)^s}, \, \, \, \, \, \, \, \, \, {\cal R}(s)>1. $

之后,形式为$ \sum\limits_{n = 0}^{\infty}\frac{a_n}{({}^{2n}_{\; n}\big) } $ ($ a_n $为$ n $的函数)的级数被广泛研究,这种级数也被称为类Apéry级数.当然关于二项式系数的另外一种级数形式$ \sum\limits_{n = 0}^{\infty}b_n\big({}^{2n}_{\; n}\big) $ ($ b_n $是$ n $的函数)也受到了很多学者的关注.利用积分、差分和特殊取值方法, Lehmer^[15]建立了许多关于这两种级数的求和公式.由Dirichlets $ L $ -级数, Sun^[19]和Zucker^[24]也建立了大量这两种类型级数的求和公式.利用Fa$ \grave{a} $ di Bruno公式, Wang和Xu^[21]建立了一系列关于调和数和二项式系数的恒等式.另外,计算机代数方法也被应用到证明和计算这些求和公式中来.例如, Ablinger^[1]提供了Mathematica包{HarmonicSums},发现并验证了大量包含中心二项式系数和调和数的无穷级数公式.

最近, Chu^[8]将超几何方法^[7]应用到建立类Apéry级数恒等式的研究中来.这种方法的应用,我们也可参见文献[5-7, 12-13, 20, 23].本文将推广超几何方法,建立更多关于中心二项式系数和广义调和数的无穷限求和公式.

广义调和数$ H_n^{(s)}(x) $定义如下

$ H_0^{(s)}(x) = 0, \, \, \, \, \, \, H_n^{(s)}(x) = \sum\limits_{j = 0}^{n-1}\frac{1}{(j+x)^s}, \, \, \, \, \, \, x\in{\Bbb C}/{\Bbb Z}^-, \, \, n, s = 1, 2, \cdots. $

当$ x = 1 $时,

$ H_0^{(s)} = 0, \, \, \, \, \, \, H_n^{(s)} = \sum\limits_{j = 1}^n\frac{1}{j^s}, \, \, \, \, \, \, n, s = 1, 2, \cdots. $

当$ s = 1 $时,

$ H_0(x) = 0, \, \, \, \, \, \, H_n(x) = \sum\limits_{j = 0}^{n-1}\frac{1}{j+x}, \, \, \, \, \, \, x\in{\Bbb C}/{\Bbb Z}^-, \, \, n = 1, 2, \cdots. $

显然,当$ x = 1 $且$ s = 1 $时, $ H_n^{(s)}(x) $即为经典调和数$ H_n $.再介绍另外一种调和数的推广形式^[9]

$ O_0^{l} = 0, \, \, \, \, \, \, O_n^{(l)} = \sum\limits_{k = 1}^{n}\frac{1}{(2k-1)^l} = \frac{1}{2^l}H_n^{(l)}(\frac{1}{2}), \, \, \, \, \, \, n, l\geq1, $

这里$ O_n = O_n^{(1)} = \sum\limits_{k = 1}^{n}\frac{1}{2k-1} $.

为了推广超几何方法,需要将Bell多项式的相关内容作一介绍.对一个序列$ t: = (t_1, t_2, \cdots) $, Bell多项式$ \Omega_i(t): = \Omega_i(t_1, t_2, \cdots, t_i) $定义为

(1.1) $ \begin{equation} \sum\limits_{m = 0}^{\infty}\Omega_m(t)x^m = \exp \Big\{\sum\limits_{k = 1}^{\infty}t_k\frac{x^k}{k}\Big\}, \end{equation} $

具体表达式为

$ \Omega_i(t) = \sum\limits_{\sigma(i)}\frac{1}{a_1!a_2!\cdots a_i!}\Big(\frac{t_1}{1}\Big)^{a_1}\Big(\frac{t_2}{2}\Big)^{a_2}\cdots\Big(\frac{t_i}{i}\Big)^{a_i}, $

这里的$ \sigma(i) $为取遍所有非负整数$ a_1, a_2, \cdots, a_i $,使得$ a_1+2a_2+\cdots+ia_i = i $,或者等价地, $ \sigma(i) $为取遍$ i $的所有分拆,参考文献[10, Sect 3.3]和[16, Sect 4.2].由上述式子,下面列出几个简单的Bell多项式的表达式:

$ \Omega_0(t) = 1, \, \, \, \, \, \Omega_1(t) = t_1, \, \, \, \, \, \Omega_2(t) = \frac{1}{2}(t_1^2+t_2), $

$ \Omega_3(t) = \frac{1}{6}(t_1^3+3t_1t_2+2t_3), \, \, \, \Omega_4(t) = \frac{1}{24}(t_1^4+6t_1^2t_2+3t_2^2+8t_1t_3+6t_4). $

令$ [x^m]f(x) $为形式幂级数$ f(x) $的$ x^m $前的系数.由生成函数方法,发现下列两个关于Bell多项式、升阶乘和广义调和数的关系式,这两个关系式分别是文章^[5]中的公式(5a)和(5b)的推广.

引理1.1  设$ \lambda $是一个变量,则有下列关系式成立:

(1.2) $ \begin{eqnarray} &&[x^m]\frac{(a+\lambda x)_n}{(a)_n} = \Omega_m(u), \, \, \, \, \, \, \, u_i = (-1)^{i-1}\lambda^iH_n^{(i)}(a), \end{eqnarray} $

(1.3) $ \begin{eqnarray} &&[x^m]\frac{(a)_n}{(a+\lambda x)_n} = \Omega_m(v), \, \, \, \, \, \, \, v_i = (-1)^{i}\lambda^iH_n^{(i)}(a), \end{eqnarray} $

其中$ (x)_n $为升阶乘,定义为

$ (x)_0 = 1, \, \, \, \, \, \, \, \, (x)_n = x(x+1)\cdots(x+n-1), \, \, \, \, \, \, \, \, \, n\in{\Bbb N}. $

证  根据对数函数的幂级数展开公式,有

$ \begin{eqnarray*} \ln\frac{(a+\lambda x)_n}{(a)_n}& = &\sum\limits_{k = 0}^{n-1}\ln(1+\frac{\lambda x}{a+k}) = \sum\limits_{k = 0}^{n-1}\sum\limits_{i = 1}^{\infty}\frac{(-1)^{i-1}}{i}\frac{(\lambda x)^i}{(a+k)^i}\\ & = &\sum\limits_{i = 1}^{\infty}\frac{(-1)^{i-1}(\lambda x)^i}{i}\sum\limits_{k = 0}^{n-1}\frac{1}{(a+k)^i} = \sum\limits_{i = 1}^{\infty}(-1)^{i-1}\lambda^iH_n^{(i)}(a)\frac{x^i}{i}. \end{eqnarray*} $

由Bell多项式的定义(1.1),公式(1.2)得证.用相似的方法,可以推导出公式(1.3).

下面我们给出gamma函数的幂级数展开公式.为此,将Weierstrass定义的gamma公式先列出来:

(1.4) $ \begin{equation} \Gamma(z) = \frac{e^{-\gamma z}}{z}\prod\limits_{n = 1}^{\infty}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}, \end{equation} $

这里$ \gamma \approx 0.577216 $是Euler-Mascheroni常数.

引理1.2  设$ \lambda $为一变量,则下列公式成立:

(1.5) $ \begin{eqnarray} \frac{\Gamma(a+x)}{\Gamma(a)} = \exp\Big\{\sum\limits_{k = 1}^{\infty}v_k(a)\frac{x^k}{k}\Big\}, \end{eqnarray} $

其中序列$ \{v_k(a)\} $定义为

$ v_1(a) = -\gamma+\sum\limits_{n = 1}^{\infty}(\frac{1}{n}-\frac{1}{a+n})-\frac{1}{a}, \, \, \, \, \, \, \, v_k(a) = (-1)^k\zeta(k, a), \, \, \, k = 2, 3, \cdots. $

证  由(1.4)式,有

$ \begin{eqnarray*} \ln\frac{\Gamma(a+x)}{\Gamma(a)}& = &\ln\Big[e^{-\gamma x}\cdot\frac{a}{a+x}\cdot\prod\limits_{n = 1}^{\infty}\big(1+\frac{x}{n+a}\big)^{-1}\cdot e^{x/n}\Big]\\ & = &-\gamma x-\ln(1+\frac{x}{a})-\sum\limits_{n = 1}^{\infty}\big[\ln(1+\frac{x}{a+n})-\frac{x}{n}\big]\\ & = &-\gamma x+\sum\limits_{k = 1}^{\infty}\frac{(-1)^k}{k}\big(\frac{x}{a}\big)^k +\sum\limits_{n = 1}^{\infty}\Big[\sum\limits_{k = 1}^{\infty}\frac{(-1)^{k}}{k}\big(\frac{x}{a+n}\big)^k+\frac{x}{n}\Big]. \end{eqnarray*} $

因此,公式(1.5)得证.

在公式(1.5)中,分别令$ a = 1 $和$ a = \frac{1}{2} $,将变量$ x $换为$ -x $,公式(1.5)将简化为文章[7]中的(0.6a)和(0.6b)式.

在经典超几何级数理论中, Gauss定理^{[3, 18]}发挥了非常重要的作用,形式如下:

(2.1) $ \begin{eqnarray} \sum\limits_{n = 0}^{\infty}\frac{(a)_n(b)_n}{n!(c)_n} = \frac{\Gamma{(c)}\Gamma{(c-a-b)}}{\Gamma{(c-a)}\Gamma{(c-b)}}, \, \, \, \, {\cal R}(c-a-b)>0. \end{eqnarray} $

在(2.1)中,作参数变换$ a\rightarrow a+\lambda x $, $ b\rightarrow b+\mu x $且$ c\rightarrow c+\sigma x $, Gauss定理转换为下列形式的无穷和恒等式:

(2.2) $ \begin{eqnarray} \sum\limits_{n = 0}^{\infty}\frac{(a)_n(b)_n}{n!(c)_n}\frac{\frac{(a+\lambda x)_n}{(a)_n} \frac{(b+\mu x)_n}{(b)_n}}{\frac{(c+\sigma x)_n}{(c)_n}} = \frac{\frac{\Gamma(c+\sigma x)}{\Gamma(c)}\frac{\Gamma(c-a-b+(\sigma-\lambda-\mu) x)} {\Gamma(c-a-b)}}{\frac{\Gamma(c-a+(\sigma-\lambda) x)} {\Gamma(c-a)}\frac{\Gamma(c-b+(\sigma-\mu) x)}{\Gamma(c-b)}} \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}. \end{eqnarray} $

提取上述式子左右两边的$ x^m $的系数,利用引理1.1和1.2,推导出如下一般求和定理.

定理2.1  对$ m\in{\Bbb N} $, $ a $, $ b $和$ c $为非负整数,并且$ c-a-b > 0 $,则

(2.3) $ \begin{eqnarray} \sum\limits_{n = 0}^{\infty}\frac{(a)_n(b)_n}{n!(c)_n}\Omega_m(r) = \frac{\Gamma{(c)}\Gamma{(c-a-b)}} {\Gamma{(c-a)}\Gamma{(c-b)}}\Omega_m(s), \end{eqnarray} $

其中两个序列$ \{r, s\} $定义为

$ r_i = (-1)^i(\sigma^iH_n^{(i)}(c)-\lambda^iH_n^{(i)}(a)-\mu^iH_n^{(i)}(b)), $

$ s_i = \sigma^iv_i(c)+(\sigma-\lambda-\mu)^iv_i(c-a-b)-(\sigma-\lambda)^iv_i(c-a)-(\sigma-\mu)^iv_i(c-b), $

这里$ v_i(x) $为在引理1.2中的表达式.

注意到,这一定理等价于文献[14]中的结果(24)和(25).当$ a $, $ b $, $ c $选取不同的特殊值时,将得到几个关于中心二项式系数的一般公式.例如,在(2.3)式中,取$ a = 1 $, $ b = \frac{1}{2} $, $ c = 2 $,有如下推论.

推论2.1  对$ m\in{\Bbb N} $,下列关于中心二项式系数的求和公式成立:

(2.4) $ \begin{equation} \sum\limits_{n = 0}^{\infty}\frac{({}^{2n}_{\;n}\big) }{4^n(n+1)}\Omega_m(r) = 2\Omega_m(s), \end{equation} $

其中

$ r_i = (-1)^i(\sigma^i(H_{n+1}^{(i)}-1)-\lambda^iH_n^{(i)}-\mu^i2^iO_n^{(i)}), $

$ s_i = \sigma^iv_i(2)+(\sigma-\lambda-\mu)^iv_i(\frac{1}{2})-(\sigma-\lambda)^iv_i(1)-(\sigma-\mu)^iv_i(\frac{3}{2}). $

根据Riemann和Hurwitz zeta函数的相关性质, $ v_i(1) $, $ v_i(2) $, $ v_i(\frac{1}{2}) $和$ v_i(\frac{3}{2}) $可表示为

$ v_i(1) = \left\{ \begin{array}{ll} -\gamma\, , &i = 1\, , \\ (-1)^i\zeta(i)\, , &i>1\, , \end{array} \right.~~ v_i(2) = \left\{ \begin{array}{ll} -\gamma+1\, , &i = 1\, , \\ (-1)^i(\zeta(i)-1)\, , &i>1\, , \end{array} \right. $

$ v_i(\frac{1}{2}) = \left\{ \begin{array}{ll} -\gamma-2\ln2\, , &i = 1\, , \\ (-1)^i(2^i-1)\zeta(i)\, , &i>1\, , \end{array} \right.~~ v_i(\frac{3}{2}) = \left\{ \begin{array}{ll} -\gamma-2\ln2+2\, , &i = 1\, , \\ (-1)^i(2^i-1)\zeta(i)-(-2)^i\, , &i>1\, . \end{array} \right. $

在(2.3)式中,令$ a = \frac{1}{2} $, $ b = \frac{1}{2} $, $ c = \frac{3}{2} $,得

推论2.2  对$ m\in{\Bbb N} $,下列关于中心二项式系数的求和公式成立:

(2.5) $ \begin{equation} \sum\limits_{n = 0}^{\infty}\frac{({}^{2n}_{\;n}\big) }{4^n(2n+1)}\Omega_m(r) = \frac{\pi}{2}\Omega_m(s), \end{equation} $

其中$ r_i = (-2\sigma)^i(O_{n+1}^{(i)}-1)-(-2)^i(\lambda^i+\mu^i)O_n^{(i)}, $

$ s_i = \sigma^iv_i(\frac{3}{2})+(\sigma-\lambda-\mu)^iv_i(\frac{1}{2})-(\sigma-\lambda)^iv_i(1)-(\sigma-\mu)^iv_i(1). $

在(2.3)式中,令$ a = \frac{1}{2} $, $ b = \frac{1}{2} $, $ c = 2 $,得

推论2.3  对$ m\in{\Bbb N} $,下列关于中心二项式系数的求和公式成立:

(2.6) $ \begin{equation} \sum\limits_{n = 0}^{\infty}\frac{{({}^{2n}_{\;n}\big) }^2}{16^n(n+1)}\Omega_m(r) = \frac{4}{\pi}\Omega_m(s), \end{equation} $

其中$ r_i = (-1)^i[\sigma^i(H_{n+1}^{(i)}-1)-(\lambda^i+\mu^i)2^iO_n^{(i)}], $

$ s_i = \sigma^iv_i(2)+(\sigma-\lambda-\mu)^iv_i(1)-(\sigma-\lambda)^iv_i(\frac{3}{2})-(\sigma-\mu)^iv_i(\frac{3}{2}). $

由上述三个推论,可以推导出一系列的关于二项式系数与广义调和数的求和公式,这些公式用下文中的表 1, 表 2, 表 3表示.

注2.1  在推论2.3和表 3中的关于中心二项式系数平方的公式在文章[4]和[22]中已被验证.

Gauss第二定理,参见文献[3, 18]:

(3.1) $ \begin{eqnarray} \sum\limits_{n = 0}^{\infty}\frac{(a)_n(c)_n}{n!(\frac{1+a+c}{2})_n}\big(\frac{1}{2}\big)^n = \frac{\Gamma{(\frac{1}{2})}\Gamma{(\frac{1+a+c}{2})}}{\Gamma{(\frac{1+a}{2})}\Gamma{(\frac{1+c}{2})}}. \end{eqnarray} $

在(3.1)式中,作变量替换$ a\rightarrow a+\lambda x $, $ c\rightarrow c+\mu x $,提取上式左右两边$ x^m $前的系数,利用引理1.1和引理1.2,可建立下列恒等式.

定理3.1  对$ m\in{\Bbb N} $, $ a $, $ c $为非负整数,下列无穷级数恒等式成立:

(3.2) $ \begin{eqnarray} \sum\limits_{n = 0}^{\infty}\frac{(a)_n(c)_n}{n!(\frac{1+a+c}{2})_n}\big(\frac{1}{2}\big)^n\Omega_m(u) = \frac{\sqrt{\pi}\Gamma{(\frac{1+a+c}{2})}}{\Gamma{(\frac{1+a}{2})}\Gamma{(\frac{1+c}{2})}}\Omega_m(t), \end{eqnarray} $

其中两个序列$ \{u, t\} $定义为:

$ u_i = (-1)^{i-1}\big[\lambda^iH_n^{(i)}(a)+\mu^iH_n^{(i)}(c)-(\frac{\lambda+\mu}{2})^iH_n^{(i)}(\frac{1+a+c}{2})\big], $

$ t_i = (\frac{\lambda+\mu}{2})^iv_i(\frac{1+a+c}{2})-(\frac{\lambda}{2})^iv_i(\frac{1+a}{2})-(\frac{\mu}{2})^iv_i(\frac{1+c}{2}). $

在(3.2)式中,令$ a = c = 1 $, $ \lambda = 1 $, $ \mu = -1 $,可推导出类Apéry级数恒等式[9, Eq 3b].

在(3.2)式中,令$ a = \frac{1}{2} $, $ c = \frac{3}{2} $,有下列推论.

推论3.1  对$ m\in{\Bbb N} $,有

(3.3) $ \begin{equation} \sum\limits_{n = 0}^{\infty}\frac{\big({}^{2n}_{\;n}\big) }{8^n}\Omega_m(u) = \sqrt{2}\Omega_m(t), \end{equation} $

其中

$ u_i = (-1)^{i-1}2^i\big[\lambda^iO_n^{(i)}+\mu^i(O_{n+1}^{(i)}-1)-(\frac{\lambda+\mu}{2})^i(O_{n+1}^{(i)}-1)\big], $

$ t_i = (\frac{\lambda+\mu}{2})^iv_i(\frac{3}{2})-(\frac{\lambda}{2})^iv_i(\frac{3}{4})-(\frac{\mu}{2})^iv_i(\frac{5}{4}). $

例3.1  在(3.3)式中,令$ m = 0 $,有

$ \sum\limits_{n = 0}^{\infty}\frac{\big({}^{2n}_{\;n}\big) }{8^n} = \sqrt{2}. $

这个恒等式在文献[15]也可看到.

例3.2  在(3.3)式中,令$ m = 1 $, $ \lambda = \mu = 1 $,有

$ \sum\limits_{n = 0}^{\infty}\frac{\big({}^{2n}_{\;n}\big)}{8^n}O_n = \frac{\sqrt{2}}{2}\ln2. $

证  在(3.3)式中,令$ m = 1 $, $ \lambda = \mu = 1 $,有

$ \begin{eqnarray*} \sum\limits_{n = 0}^{\infty}\frac{\big({}^{2n}_{\;n}\big)}{8^n}O_n = \frac{\sqrt{2}}{2}t_1, \, \, \, \, \, \, t_1 = -\zeta(1, \frac{3}{2})+\frac{1}{2}\zeta(1, \frac{3}{4}) +\frac{1}{2}\zeta(1, \frac{5}{4}). \end{eqnarray*} $

Hurwitz zeta函数($ m $是非负整数)与polygamma函数的关系为

(3.4) $ \begin{eqnarray} \psi^{(m)}(z) = (-1)^{m+1}m!\zeta(m+1, z). \end{eqnarray} $

由这个关系式,有

$ t_1 = \psi(\frac{3}{2})-\frac{1}{2}\psi(\frac{3}{4})-\frac{1}{2}\psi(\frac{5}{4}), $

其中$ \psi(x) = \psi^{(0)}(x) $是digamma函数,关于它的一些特殊值可以参考文献[17]:

$ \begin{eqnarray*} \psi(\frac{1}{2}) = -2\ln2-\gamma, \, \, \, \, \psi(\frac{1}{4}) = -\frac{\pi}{2}-3\ln2-\gamma. \end{eqnarray*} $

根据digamma函数的关系式:

$ \psi(x+1) = \psi(x)+\frac{1}{x}, $

$ \psi(1-x)-\psi(x) = \pi\cot(\pi x), $

可以得到

$ \begin{eqnarray*} \psi(\frac{3}{2}) = -2\ln2-\gamma+2, \, \, \, \, \psi(\frac{3}{4}) = -\frac{\pi}{2}-3\ln2-\gamma+\pi, \, \, \, \, \psi(\frac{5}{4}) = -\frac{\pi}{2}-3\ln2-\gamma+4. \end{eqnarray*} $

因此$ t_1 = \ln2 $,进而结论得证.

相似地,有下列求和公式成立.

例3.3  在(3.3)式中,令$ m = 1 $, $ \lambda = 0 $, $ \mu = 1 $,有

$ \sum\limits_{n = 0}^{\infty}\frac{\big({}^{2n}_{\;n}\big) }{8^n}O_{n+1} = \sqrt{2}(\frac{1}{2}\ln2+\frac{\pi}{4}). $

例3.4  在(3.3)式中,令$ m = 2 $, $ \lambda = \mu = 1 $,有

$ \sum\limits_{n = 0}^{\infty}\frac{\big({}^{2n}_{\;n}\big) }{8^n}(O_{n}^2-O_n^{(2)}) = \frac{\sqrt{2}}{4}\ln^22. $

证  在(3.3)式中,令$ m = 2 $, $ \lambda = \mu = 1 $,利用关系式(3.4),有

$ \begin{eqnarray*} \sum\limits_{n = 0}^{\infty}\frac{\big({}^{2n}_{\;n}\big) }{8^n}(O_{n}^2-O_n^{(2)}) = \frac{\sqrt{2}}{4}(t_1^2+t_2), \, \, \, \, \, \, t_2 = \psi'(\frac{3}{2})-\frac{1}{4}\psi'(\frac{3}{4}) -\frac{1}{4}\psi'(\frac{5}{4})), \end{eqnarray*} $

其中$ \psi'(x) = \psi^{(1)}(x) $.而$ \psi'(\frac{1}{2}) $, $ \psi'(\frac{1}{4}) $以及$ \psi'(\frac{3}{4}) $的值为

$ \begin{eqnarray*} \psi'(\frac{1}{2}) = \frac{\pi^2}{2}, \, \, \, \, \psi'(\frac{1}{4}) = \pi^2+8G, \, \, \, \psi'(\frac{3}{4}) = \pi^2-8G, \end{eqnarray*} $

其中$ G $为Catalan常数,这些值可参考文献[11].另外,由polygamma函数的递推关系

(3.5) $ \begin{equation} \psi^{(n)}(x+1) = \psi^{(n)}(x)+\frac{(-1)^nn!}{x^{n+1}}, \end{equation} $

可得$ \psi'(\frac{3}{2}) = \frac{\pi^2}{2}-4 $, $ \psi'(\frac{5}{4}) = \pi^2+8G-16 $,因此计算出$ t_2 = 0 $.结论得证.

例3.5  在(3.3)式中,令$ m = 2 $, $ \lambda = 0 $, $ \mu = 1 $,有

$ \sum\limits_{n = 0}^{\infty}\frac{\big({}^{2n}_{\;n}\big) }{8^n}(O_{n+1}^2-2O_{n+1}-3O_{n+1}^{(2)}) = \sqrt{2}(\frac{1}{4}\ln^22+\frac{\pi}{4}\ln2 -\ln2-\frac{\pi}{2}-\frac{\pi^2}{16}-2G). $

例3.6  在(3.3)式中,令$ m = 3 $, $ \lambda = \mu = 1 $,有

$ \sum\limits_{n = 0}^{\infty}\frac{\big({}^{2n}_{\;n}\big) }{8^n}(O_{n}^3-3O_nO_n^{(2)}+2O_n^{(3)}) = \frac{\sqrt{2}}{8}\ln^32. $

证  在(3.3)式中,令$ m = 3 $, $ \lambda = \mu = 1 $,利用关系式(3.4),有

$ \sum\limits_{n = 0}^{\infty}\frac{\big({}^{2n}_{\;n}\big) }{8^n}(O_{n}^3-3O_nO_n^{(2)}+2O_n^{(3)}) = \frac{\sqrt{2}}{8}(t_1^3+3t_1t_2+2t_3), $

$ t_3 = \frac{1}{2}(\psi''(\frac{3}{2})-\frac{1}{8}\psi''(\frac{3}{4})-\frac{1}{8}\psi''(\frac{5}{4})), $

其中$ \psi''(x) = \psi^{(2)}(x) $.在文献[17]中,发现

$ \begin{eqnarray*} \psi''(\frac{1}{2}) = -14\zeta(3), \, \, \, \, \psi''(\frac{1}{4}) = -2\pi^3-56\zeta(3), \, \, \, \psi''(\frac{3}{4}) = 2\pi^3-56\zeta(3). \end{eqnarray*} $

根据关系式(3.5),可以推导出$ t_3 = 0 $,结论得证.

例3.7  在(3.3)式中,令$ m = 3 $, $ \lambda = 0 $, $ \mu = 1 $

$ \begin{eqnarray*} &&\sum\limits_{n = 0}^{\infty}\frac{\big({}^{2n}_{\;n}\big) }{8^n}(O_{n+1}^3-3O_{n+1}^2+12O_{n+1}-9O_{n+1}O_{n+1}^{(2)}+9O_{n+1}^{(2)}+14O_{n+1}^{(3)})\\ & = &\sqrt{2}(6\ln2-\frac{3}{4}\ln^22+\frac{\ln^32}{8}+3\pi-\frac{3}{4}\ln2\cdot\pi+\frac{3}{16}\ln^22\cdot\pi+\frac{3}{16}\pi^2\\ &&-\frac{3}{32}\ln2\cdot\pi^2+\frac{11}{64}\pi^3+6G-3\ln2\cdot G-\frac{3}{2}\pi G+\frac{21}{4}\zeta(3)). \end{eqnarray*} $