Rogers-Ramanujan型连分数的构造方法

The Construction Method of Rogers-Ramanujan Type Continued Fractions

摘要: Rogers-Ramanujan连分数是Ramanujan笔记本的重要研究内容之一。本文利用两个已知Rogers恒等式，使用数学归纳法和递归关系式构造了一个新的Rogers-Ramanujan型连分数。

Abstract: The Rogers-Ramanujan type continued fraction is one of the significant research contents in Ramanujan’s notebooks. In this paper, by using two known Rogers’s identities, the mathematical induction and a recursive relation, a new Rogers-Ramanujan type continued fraction is constructed.

1. 前言

Rogers-Ramanujan型连分数不仅具有优美的数学结构，还与超几何函数、椭圆积分、模等式等数学分支紧密相关，在数论、统计学、代数学以及组合分拆理论等领域有重要应用。发现和证明新的Rogers-Ramanujan型恒等式始终是一个非常活跃的课题。在2010年，Helmut Prodinger [1]利用递归关系：$z{S}_{k+1}=S_{k-1}-a_k{S}_k$ ，($k=0,1,2,\cdots$ )和数学归纳法对Rogers-Ramanujan型连分数进行研究。在2011年，Nancy S. S. Gu和Helmut Prodinger [2]运用同样的递归关系对大量的Rogers-Ramanujan型恒等式进行研究，得到了许多Rogers-Ramanujan型恒等式的连分数展开，总计18个结果。在2013年，Kamilla Oliver和Helmut Prodinger [3]根据小分拆定理推导出7个Rogers-Ramanujan型连分数展开式。在2023年，蔡华蓉[4]对上述结果之一进行了修正。在2025年，张英凡[5]对Rogers-Ramanujan型恒等式进行研究，并利用新的递归关系：$z{s}_{k+1}=s_{k-1}-\left(a_k+b_{k}z\right)s_k,\left(k=0,1,2,\cdots \right)$ 得到了一个Rogers-Ramanujan型恒等式的连分数展开。

本文利用两个已知Rogers恒等式[6]，运用递归关系$z{s}_{k+1}=s_{k-1}-\left(a_k+b_{k}z\right)s_k,\left(k=0,1,2,\cdots \right)$ 和数学归纳法，构造出一个Rogers-Ramanujan型连分数。

为了方便，本节对后续内容中所用到的定义做如下说明。

定义1.1

$\begin{array}{l}{\left(a;q\right)}_n:={\displaystyle \prod }_{k=0}^{n-1}\left(1-a{q}^k\right),n\ge 1,\\ {\left(a;q\right)}_{\infty }:={\displaystyle \prod }_{k=0}^{\infty }\left(1-a{q}^k\right),\left|q\right|<1.\end{array}$

特别地，当$n=0$ 时

${\left(a;q\right)}_0:=1$

2. 主要结论及证明

在本节中，首先给出以下两个恒等式：

Rogers恒等式1：

${\displaystyle \sum }_{n=0}^{\infty }\frac{q^{n^2}}{{\left(q;q\right)}_{2n}}=\frac{{\left(q^8,q^{12},q^{20};q^{20}\right)}_{\infty }{\left(-q;q^2\right)}_{\infty }}{{\left(q^2;q^2\right)}_{\infty }}.$

Rogers恒等式2：

${\displaystyle \sum }_{n=0}^{\infty }\frac{q^{n^2+2n}}{{\left(q;q\right)}_{2n+1}}=\frac{{\left(q^4,q^6,q^{10};q^{10}\right)}_{\infty }{\left(q^2,q^{18};q^{20}\right)}_{\infty }}{{\left(q;q\right)}_{\infty }}.$

引理2.1 (Guarticle 2)设

$F\left(z\right)={\displaystyle \sum }_{n=0}^{\infty }\frac{z^n{q}^{n^2+2n}}{{\left(q;q\right)}_{2n}},G\left(z\right)={\displaystyle \sum }_{n=0}^{\infty }\frac{z^n{q}^{n^2}}{{\left(q;q\right)}_{2n}}.$

则

$\begin{array}{c}\frac{zF\left(z\right)}{G\left(z\right)}=\frac{z\left(1-q\right)}{1-}\frac{zq}{\left(1-q\right)\left(1-q^3\right)+}\frac{zq\left(1-q^6\right)}{1-q^5+}\frac{z{q}^8\left(1-q\right)}{\left(1-q^6\right)\left(1-q^7\right)+}\cdots \\ \frac{z{q}^{2n-1}\left(1-q^{2n^2+3n+1}\right)}{1-q^{4n+1}+}\frac{z{q}^{6n+2}\left(1-q^{2n^2-n}\right)}{\left(1-q^{4n+3}\right)\left(1-q^{2n^2+3n+1}\right)+}\cdots \end{array}$ (1)

我们使用上述两个恒等式，首先定义出两个幂级数$F\left(z\right)$ 和$G\left(z\right)$ ，构造幂级数$\left\{s_k\right\}$ ，和数$\left\{a_k\right\},\left\{b_k\right\}$ 以满足递归关系$z{s}_{k+1}=s_{k-1}-\left(a_k+b_{k}z\right)s_k,\left(k=0,1,2,\cdots \right)$ ，由此求出幂级数$s_k$ ，和数$a_k,b_k$ ，并由数学归纳法进行证明。从而构造出新的Rogers-Ramanujan型连分数。

定理2.1 设

$F\left(z\right)={\displaystyle \sum }_{n=0}^{\infty }\frac{z^n{q}^{n^2+2n}}{{\left(q;q\right)}_{2n}},G\left(z\right)={\displaystyle \sum }_{n=0}^{\infty }\frac{z^n{q}^{n^2}}{{\left(q;q\right)}_{2n}}.$

则

$\begin{array}{c}\frac{zF\left(z\right)}{G\left(z\right)}=\frac{z}{1+qz+}\frac{z{q}^2}{1-q+q^{3}z+}\frac{z{q}^6}{1-q^3+q^{5}z+}\cdots \\ \frac{z{q}^{8k-6}}{1-q^{4k-3}+q^{4k-1}z+}\frac{z{q}^{8k-2}}{1-q^{4k-1}+q^{4k+1}z+}\cdots \end{array}$ (2)

证明 令幂级数$F\left(z\right)=s_0,G\left(z\right)=s_{-1}$ ，首先根据递归关系式$z{s}_{k+1}=s_{k-1}-\left(a_k+b_{k}z\right)s_k,\left(k=0,1,2,\cdots \right)$ 求出幂级数$\left\{s_k\right\}$ 的前七项$s_{-1},s_0,s_1,s_2,s_3,s_4,s_5$ 以及对应的数列$\left\{a_k\right\}$ 的前五项$a_0,a_1,a_2,a_3,a_4$ ，其中取数列$\left\{b_k\right\}$ 为$b_{2k}=q^{2k+1},b_{2k+1}=q^{2k+1}$ ，再猜测幂级数$\left\{s_k\right\}$ 和数$\left\{a_k\right\},\left\{b_k\right\}$ ，最后用数学归纳法证明。其中记幂级数$\left\{s_k\right\}$ 中$z^n$ 项的系数为$\left[z^n\right]s_k$

当$k=0$ 时，有$z{s}_1=s_{-1}-\left(a_0+b_{0}z\right)s_0$ ，其中取$b_0=q$ 。

对于$z^0$ 的系数有

$0=1-a_0.$

则可得

$a_0=1.$

对于$z^n$ 的系数有

$\begin{array}{c}\left[z^n\right]z{s}_1=\left[z^n\right]\left(s_{-1}-\left(a_0+b_{0}z\right)s_0\right)\\ =\frac{q^{n^2}}{{\left(q;q\right)}_{2n}}-\frac{q^{n^2+2n}}{{\left(q;q\right)}_{2n}}-q\frac{q^{{\left(n-1\right)}^2+2\left(n-1\right)}}{{\left(q;q\right)}_{2\left(n-1\right)}}\\ =\frac{q^{n^2}}{{\left(q;q\right)}_{2n-1}}-\frac{q^{n^2}}{{\left(q;q\right)}_{2n-2}}\\ =\frac{q^{n^2}}{{\left(q;q\right)}_{2n-1}}\left(1-\left(1-q^{2n-1}\right)\right)\\ =\frac{q^{n^2+2n-1}}{{\left(q;q\right)}_{2n-1}}.\end{array}$ (3)

则可得

$\left[z^n\right]s_1=\frac{q^{n^2+4n+2}}{{\left(q;q\right)}_{2n+1}}.$

即

$s_1={\displaystyle \sum }_{n=0}^{\infty }\frac{q^{n^2+4n+2}}{{\left(q;q\right)}_{2n+1}}{z}^n.$

当$k=1$ 时，有$z{s}_2=s_0-\left(a_1+b_{1}z\right)s_1$ ，其中取$b_1=q$ 。

对于$z^0$ 的系数有

$0=1-a_1\frac{q^2}{1-q}.$

则可得

$a_1=\frac{1-q}{q^2}.$

对于$z^n$ 的系数有

$\begin{array}{c}\left[z^n\right]z{s}_2=\left[z^n\right]\left(s_0-\left(a_1+b_{1}z\right)s_1\right)\\ =\frac{q^{n^2+2n}}{{\left(q;q\right)}_{2n}}-\frac{1-q}{q^2}\frac{q^{n^2+4n+2}}{{\left(q;q\right)}_{2n+1}}-q\frac{q^{{\left(n-1\right)}^2+4\left(n-1\right)+2}}{{\left(q;q\right)}_{2\left(n-1\right)+1}}\\ =\frac{q^{n^2+2n}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)}-\frac{q^{n^2+2n}}{{\left(q;q\right)}_{2n-1}}\\ =\frac{q^{n^2+2n}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)}\left(1-\left(1-q^{2n+1}\right)\right)\\ =\frac{q^{n^2+4n+1}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)}.\end{array}$ (4)

则可得

$\left[z^n\right]s_2=\frac{q^{n^2+6n+6}}{{\left(q;q\right)}_{2n+1}\left(1-q^{2n+3}\right)}.$

即

$s_2={\displaystyle \sum }_{n=0}^{\infty }\frac{q^{n^2+6n+6}}{{\left(q;q\right)}_{2n+1}\left(1-q^{2n+3}\right)}{z}^n.$

当$k=2$ 时，有$z{s}_3=s_1-\left(a_2+b_{2}z\right)s_2$ ，其中取$b_2=q$ 。

对于$z^0$ 的系数有

$0=\frac{q^2}{1-q}-a_2\frac{q^6}{\left(1-q\right)\left(1-q^3\right)}.$

则可得

$a_2=\frac{1-q^3}{q^4}.$

对于$z^n$ 的系数有

$\begin{array}{c}\left[z^n\right]z{s}_3=\left[z^n\right]\left(s_1-\left(a_2+b_{2}z\right)s_2\right)\\ =\frac{q^{n^2+4n+2}}{{\left(q;q\right)}_{2n+1}}-\frac{1-q^3}{q^4}\frac{q^{n^2+6n+6}}{{\left(q;q\right)}_{2n+1}\left(1-q^{2n+3}\right)}-q\frac{q^{{\left(n-1\right)}^2+6\left(n-1\right)+6}}{{\left(q;q\right)}_{2\left(n-1\right)+1}\left(1-q^{2\left(n-1\right)+3}\right)}\\ =\frac{q^{n^2+4n+2}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)\left(1-q^{2n+3}\right)}-\frac{q^{n^2+4n+2}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)}\\ =\frac{q^{n^2+4n+2}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)\left(1-q^{2n+3}\right)}\left(1-\left(1-q^{2n+3}\right)\right)\\ =\frac{q^{n^2+6n+5}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)\left(1-q^{2n+3}\right)}.\end{array}$ (5)

则可得

$\left[z^n\right]s_3=\frac{q^{n^2+8n+12}}{{\left(q;q\right)}_{2n+1}\left(1-q^{2n+3}\right)\left(1-q^{2n+5}\right)}.$

即

$s_3={\displaystyle \sum }_{n=0}^{\infty }\frac{q^{n^2+8n+12}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_2}{z}^n.$

当$k=3$ 时，有$z{s}_4=s_2-\left(a_3+b_{3}z\right)s_3$ ，其中取$b_3=q$ 。

对于$z^0$ 的系数有

$0=\frac{q^6}{\left(1-q\right)\left(1-q^3\right)}-a_3\frac{q^{12}}{\left(1-q\right)\left(1-q^3\right)\left(1-q^5\right)}.$

则可得

$a_3=\frac{1-q^5}{q^6}.$

对于$z^n$ 的系数有

$\begin{array}{c}\left[z^n\right]z{s}_4=\left[z^n\right]\left(s_2-\left(a_3+b_{3}z\right)s_3\right)\\ =\frac{q^{n^2+6n+6}}{{\left(q;q\right)}_{2n+1}\left(1-q^{2n+3}\right)}-\frac{1-q^5}{q^6}\frac{q^{n^2+8n+12}}{{\left(q;q\right)}_{2n+1}\left(1-q^{2n+3}\right)\left(1-q^{2n+5}\right)}\\ -q\frac{q^{{\left(n-1\right)}^2+8\left(n-1\right)+12}}{{\left(q;q\right)}_{2\left(n-1\right)+1}\left(1-q^{2\left(n-1\right)+3}\right)\left(1-q^{2\left(n-1\right)+5}\right)}\\ =\frac{q^{n^2+6n+6}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)\left(1-q^{2n+3}\right)\left(1-q^{2n+5}\right)}-\frac{q^{n^2+6n+6}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)\left(1-q^{2n+3}\right)}\end{array}$

$=\frac{q^{n^2+8n+11}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)\left(1-q^{2n+3}\right)\left(1-q^{2n+5}\right)}.$ (6)

则可得

$\left[z^n\right]s_4=\frac{q^{n^2+10n+20}}{{\left(q;q\right)}_{2n+1}\left(1-q^{2n+3}\right)\left(1-q^{2n+5}\right)\left(1-q^{2n+7}\right)}.$

即

$s_4={\displaystyle \sum }_{n=0}^{\infty }\frac{q^{n^2+10n+20}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_3}{z}^n.$

当$k=4$ 时，有$z{s}_5=s_3-\left(a_4+b_{4}z\right)s_4$ ，其中取$b_4=q$ 。

对于$z^0$ 的系数有

$0=\frac{q^{12}}{\left(1-q\right)\left(1-q^3\right)\left(1-q^5\right)}-a_4\frac{q^{20}}{\left(1-q\right)\left(1-q^3\right)\left(1-q^5\right)\left(1-q^7\right)}.$

则可得

$a_4=\frac{1-q^7}{q^8}.$

对于$z^n$ 的系数有

$\begin{array}{c}\left[z^n\right]z{s}_5=\left[z^n\right]\left(s_3-\left(a_4+b_{4}z\right)s_4\right)\\ =\frac{q^{n^2+8n+12}}{{\left(q;q\right)}_{2n+1}\left(1-q^{2n+3}\right)\left(1-q^{2n+5}\right)}\\ -\frac{1-q^7}{q^8}\frac{q^{n^2+10n+20}}{{\left(q;q\right)}_{2n+1}\left(1-q^{2n+3}\right)\left(1-q^{2n+5}\right)\left(1-q^{2n+7}\right)}\\ -q\frac{q^{{(n-1)}^2+10\left(n-1\right)+20}}{{\left(q;q\right)}_{2\left(n-1\right)+1}\left(1-q^{2\left(n-1\right)+3}\right)\left(1-q^{2\left(n-1\right)+5}\right)\left(1-q^{2\left(n-1\right)+7}\right)}\\ =\frac{q^{n^2+8n+12}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)\left(1-q^{2n+3}\right)\left(1-q^{2n+5}\right)\left(1-q^{2n+7}\right)}\\ -\frac{q^{n^2+8n+12}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)\left(1-q^{2n+3}\right)\left(1-q^{2n+5}\right)}\\ =\frac{q^{n^2+10n+19}}{{\left(q;q\right)}_{2n-1}\left(1-q^{2n+1}\right)\left(1-q^{2n+3}\right)\left(1-q^{2n+5}\right)\left(1-q^{2n+7}\right)}.\end{array}$ (7)

则可得

$\left[z^n\right]s_5=\frac{q^{n^2+12n+30}}{{\left(q;q\right)}_{2n+1}\left(1-q^{2n+3}\right)\left(1-q^{2n+5}\right)\left(1-q^{2n+7}\right)\left(1-q^{2n+9}\right)}.$

即

$s_5={\displaystyle \sum }_{n=0}^{\infty }\frac{q^{n^2+12n+30}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_4}{z}^n.$

当$k=5$ 时，有$z{s}_6=s_4-\left(a_5+b_{5}z\right)s_5$ ，其中取$b_5=q$ 。

对于$z^0$ 的系数有

$0=\frac{q^{20}}{\left(1-q\right)\left(1-q^3\right)\left(1-q^5\right)\left(1-q^7\right)}-a_5\frac{q^{30}}{\left(1-q\right)\left(1-q^3\right)\left(1-q^5\right)\left(1-q^7\right)\left(1-q^9\right)}.$

则可得

$a_5=\frac{1-q^9}{q^{10}}.$

对于$z^n$ 的系数有

$\begin{array}{c}\left[z^n\right]z{s}_6=\left[z^n\right]\left(s_4-\left(a_5+b_{5}z\right)s_5\right)\\ =\frac{q^{n^2+10n+20}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_3}-\frac{1-q^9}{q^{10}}\frac{q^{n^2+12n+30}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_4}\\ -q\frac{q^{{\left(n-1\right)}^2+12\left(n-1\right)+30}}{{\left(q;q\right)}_{2\left(n-1\right)+1}{\left(q^{2\left(n-1\right)+3};q^2\right)}_4}\\ =\frac{q^{n^2+10n+20}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_5}-\frac{q^{n^2+10n+20}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_4}\\ =\frac{q^{n^2+10n+20}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_5}\left(1-\left(1-q^{2n+9}\right)\right)\\ =\frac{q^{n^2+12n+29}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_5}.\end{array}$ (8)

则可得

$\left[z^n\right]s_6=\frac{q^{n^2+14n+42}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_5}.$

即

$s_6={\displaystyle \sum }_{n=0}^{\infty }\frac{q^{n^2+14n+42}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_5}{z}^n.$

综上可猜测辅助级数$s_k$ 和数$a_k$ ，$b_k$

$\begin{array}{l}{s}_{2k}={\displaystyle \sum }_{n=0}^{\infty }\frac{q^{n^2+\left(4k+2\right)n+\left(4k^2+2k\right)}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_{2k-1}}{z}^n.\\ s_{2k+1}={\displaystyle \sum }_{n=0}^{\infty }\frac{q^{n^2+\left(4k+4\right)n+\left(4k^2+6k+2\right)}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_{2k}}{z}^n.\end{array}$

$\begin{array}{l}{a}_0=1,a_{2k}=\frac{1-q^{4k-1}}{q^{4k}},a_{2k+1}=\frac{1-q^{4k+1}}{q^{4k+2}}.\\ b_{2k}=q,b_{2k+1}=q.\end{array}$ (9)

接下来证明猜测的辅助级数$s_k$ 和数$a_k$ ，$b_k$ 满足递归关系式$z{s}_{k+1}=s_{k-1}-\left(a_k+b_{k}z\right)s_k,\left(k=0,1,2,\cdots \right)$ ，即证

$\left[z^n\right]z{s}_{2k+2}=\left[z^n\right]\left(s_{2k}-\left(a_{2k+1}+b_{2k+1}z\right)s_{2k+1}\right).$

$\left[z^n\right]z{s}_{2k+1}=\left[z^n\right]\left(s_{2k-1}-\left(a_{2k}+b_{2k}z\right)s_{2k}\right).$

对于$\left[z^n\right]z{s}_{2k+2}=\left[z^n\right]\left(s_{2k}-\left(a_{2k+1}+b_{2k+1}z\right)s_{2k+1}\right)$ 。当$k=0$ 时，等式成立。

当$k>0$ 时，

$\left[z^n\right]s_{2k+2}=\frac{q^{n^2+\left(4k+6\right)n+\left(4k^2+10k+6\right)}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_{2k+1}}.$

$\begin{array}{c}\left[z^n\right]z{s}_{2k+2}=\frac{q^{{\left(n-1\right)}^2+\left(4k+6\right)\left(n-1\right)+\left(4k^2+10k+6\right)}}{{\left(q;q\right)}_{2\left(n-1\right)+1}{\left(q^{2\left(n-1\right)+3};q^2\right)}_{2k+1}}\\ =\frac{q^{n^2+\left(4k+4\right)n+\left(4k^2+6k+1\right)}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_{2k+1}}.\end{array}$ (10)

$\begin{array}{l}\left[z^n\right]\left(s_{2k}-\left(a_{2k+1}+b_{2k+1}z\right)s_{2k+1}\right)\\ =\frac{q^{n^2+\left(4k+2\right)n+\left(4k^2+2k\right)}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_{2k-1}}-\frac{1-q^{4k+1}}{q^{4k+2}}\frac{q^{n^2+\left(4k+4\right)n+\left(4k^2+6k+2\right)}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_{2k}}\\ -q\frac{q^{{(n-1)}^2+\left(4k+4\right)\left(n-1\right)+\left(4k^2+6k+2\right)}}{{\left(q;q\right)}_{2\left(n-1\right)+1}{\left(q^{2\left(n-1\right)+3};q^2\right)}_{2k}}\\ =\frac{q^{n^2+\left(4k+2\right)n+\left(4k^2+2k\right)}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_{2k+1}}-\frac{q^{n^2+\left(4k+2\right)n+\left(4k^2+2k\right)}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_{2k}}\\ =\frac{q^{n^2+\left(4k+2\right)n+\left(4k^2+2k\right)}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_{2k+1}}\left(1-\left(1-q^{2n+4k+1}\right)\right)\\ =\frac{q^{n^2+\left(4k+4\right)n+\left(4k^2+6k+1\right)}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_{2k+1}}\end{array}$ (11)

对于$\left[z^n\left]z{s}_{2k+1}=\right[z^n\right]\left(s_{2k-1}-\left(a_{2k}+b_{2k}z\right)s_{2k}\right)$ 。当$k=0$ 时，等式成立。

当$k>0$ 时，

$\left[z^n\right]s_{2k-1}=\frac{q^{n^2+4kn+4k^2-2k}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_{2k-2}}.$

$\left[z^n\right]\left(s_{2k-1}-\left(a_{2k}+b_{2k}z\right)s_{2k}\right)$

$\begin{array}{l}=\frac{q^{n^2+4kn+4k^2-2k}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_{2k-2}}-\frac{1-q^{4k-1}}{q^{4k}}\frac{q^{n^2+\left(4k+2\right)n+\left(4k^2+2k\right)}}{{\left(q;q\right)}_{2n+1}{\left(q^{2n+3};q^2\right)}_{2k-1}}\\ -q\frac{q^{{\left(n-1\right)}^2+\left(4k+2\right)\left(n-1\right)+\left(4k^2+2k\right)}}{{\left(q;q\right)}_{2\left(n-1\right)+1}{\left(q^{2\left(n-1\right)+3};q^2\right)}_{2k-1}}\\ =\frac{q^{n^2+4kn+4k^2-2k}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_{2k}}-\frac{q^{n^2+4kn+4k^2-2k}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_{2k-1}}\\ =\frac{q^{n^2+4kn+4k^2-2k}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_{2k}}\left(1-\left(1-q^{2n+4k-1}\right)\right)\\ =\frac{q^{n^2+\left(4k+2\right)n+4k^2+2k-1}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_{2k}}.\end{array}$ (12)

$\begin{array}{c}\left[z^n\right]z{s}_{2k+1}=\frac{q^{{\left(n-1\right)}^2+4\left(k+1\right)\left(n-1\right)+\left(4k^2+6k+2\right)}}{{\left(q;q\right)}_{2\left(n-1\right)+1}{\left(q^{2\left(n-1\right)+3};q^2\right)}_{2k}}\\ =\frac{q^{n^2+\left(4k+2\right)n+4k^2+2k-1}}{{\left(q;q\right)}_{2n-1}{\left(q^{2n+1};q^2\right)}_{2k}}.\end{array}$ (13)

则上述递归关系式成立。

将$a_k$ ，$b_k$ 代入得

$\frac{zF\left(z\right)}{G\left(z\right)}=\frac{z}{1+qz+}\frac{z}{\frac{1-q}{q^2}+qz+}\frac{z}{\frac{1-q^3}{q^4}+qz+}\cdots \frac{z}{\frac{1-q^{4k-3}}{q^{4k-2}}+qz+}\frac{z}{\frac{1-q^{4k-1}}{q^{4k}}+qz+}\cdots .$

化简得

$\frac{zF\left(z\right)}{G\left(z\right)}=\frac{z}{1+qz+}\frac{z{q}^2}{1-q+q^{3}z+}\frac{z{q}^6}{1-q^3+q^{5}z+}\cdots \frac{z{q}^{8k-6}}{1-q^{4k-3}+q^{4k-1}z+}\frac{z{q}^{8k-2}}{1-q^{4k-1}+q^{4k+1}z+}\cdots .$

证毕。

特别地，在定理2.1中，取$z=1$ 可得

$\begin{array}{c}\frac{F\left(1\right)}{G\left(1\right)}=\frac{{\displaystyle \sum _{n=0}^{\infty }\frac{q^{n^2+2n}}{{\left(q;q\right)}_{2n}}}}{{\displaystyle \sum _{n=0}^{\infty }\frac{q^{n^2}}{{\left(q;q\right)}_{2n}}}}\\ =\frac{1}{1+q+}\frac{q^2}{1-q+q^3+}\frac{q^6}{1-q^3+q^5+}\cdots \\ \frac{q^{8k-6}}{1-q^{4k-3}+q^{4k-1}+}\frac{q^{8k-2}}{1-q^{4k-1}+q^{4k+1}+}\cdots .\end{array}$ (14)

注：以上结果在其他文献中并无发现，故为新的连分数。