Summation Formulae for Quintic q-Series

Abstract

By utilizing the modified Abel lemma on summation by parts, we examine a class of quintic q-series, that have close connections to the “twisted cubic q-series”. Several remarkable summation and transformation formulae are established. The related reversal series are also reviewed briefly.

1. Introduction and Motivation

Let $\mathbb{N}$ be the set of natural numbers with ${\mathbb{N}}_0=\left\{0\right\}\cup \mathbb{N}$. For an indeterminate x, the rising factorial with the base q is defined by ${(x;q)}_0=1$ and

$${(x;q)}_n=(1-x)(1-qx)\cdots (1-q^{n-1}x)\mathrm{for}n\in \mathbb{N}.$$

Its multiparameter forms will be abbreviated as follows:

$$\begin{array}{cc}\hfill {\left[A,B,\cdots ,C;q\right]}_n& ={(A;q)}_n{(B;q)}_n\cdots {(C;q)}_n,\hfill \\ \hfill {\left[\begin{array}{c}\alpha ,\beta ,\cdots ,\gamma \\ A,B,\cdots ,C\end{array}|q\right]}_n& =\frac{{\left(\alpha ;q\right)}_n{\left(\beta ;q\right)}_n\cdots {\left(\gamma ;q\right)}_n}{{\left(A;q\right)}_n{\left(B;q\right)}_n\cdots {\left(C;q\right)}_n}.\hfill \end{array}$$

The basic hypergeometric series (shortly “q-series”, cf. [1,2,3]) playsplay an important role in pure and applied sciences [4,5,6,7,8,9] for their wide applications to number theory [2,10,11], theoretical physics [12,13] and combinatorial enumeration [12,14]. During the past few decades, there has been growing interest in finding new identities and transformation formulae. In particular, there was an intensive exploration to well–poised series [15,16,17,18], quadratic series [19,20,21,22], cubic series [19,21,23,24], and quartic series [19,20,25,26], mainly made by Gasper and Rahman (see also [2], §3.8) and the author of this paper, with his collaborators.

A quarter of a century ago, the author ([27], Equations 5.4d: $\epsilon =1,2$) found, by making use of inversion techniques, the following two summation formulae

$$\begin{array}{cc}\hfill \sum _{k=0}^n& \left(1-q^{6k+1/2}\right)\frac{{[q^{-5n},q^{5n+5/2};q^5]}_k}{{[q^{-1-5n},q^{5n+3/2};q]}_k}\frac{{[q^{3/2},q^{-1/2};q]}_{2k}}{{(q^2;q)}_{4k}}{q}^k=\frac{{(q^5;q^5)}_n{(q^{\frac{1}{2}};q)}_{5n+1}}{{(q^{\frac{5}{2}};q^5)}_n{(q^2;q)}_{5n}},\hfill \\ \hfill \sum _{k=0}^n& \left(1-q^{6k+7/2}\right)\frac{{[q^{-5n},q^{5n+3/2};q^5]}_k}{{[q^{-3-5n},q^{5n+9/2};q]}_k}\frac{{[q^{5/2},q^{1/2};q]}_{2k}}{{(q^4;q)}_{4k}}{q}^k=\frac{{(q^5;q^5)}_n{(q^{\frac{5}{2}};q)}_{5n+2}}{{(q^{\frac{5}{2}};q^5)}_{n+1}{(q^4;q)}_{5n}}.\hfill \end{array}$$

They are isolated examples of the “quintic series” defined by the partial sums

$$\begin{array}{cc}\hfill {\Lambda }_n(a,c)=& \sum _{k=0}^n\left(1-q^{6k-1}{a}^3\right)\frac{{[a^{2}c,a^3/c;q^5]}_k}{{[c,a/c;q]}_k}\frac{{[qa,a/q;q]}_{2k}}{{(q{a}^2;q)}_{4k}}{q}^k,\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill {\lambda }_n(a,c)=& \sum _{k=0}^n\left(1-q^{6k+1}{a}^3\right)\frac{{[qc,qa/c;q]}_k}{{[q^5{a}^{2}c,q^5{a}^3/c;q^5]}_k}\frac{{(a^2;q)}_{4k}}{{[a,q^{2}a;q]}_{2k}}{q}^k;\hfill \end{array}$$

(2)

which satisfy obviously the following properties:

$${\Lambda }_n(a,c)={\Lambda }_n(a,a/c)\mathrm{and}{\lambda }_n(a,c)={\lambda }_n(a,a/c).$$

These two series are called “quintic series” since the factorials of order k appearing in the numerator and the denominator can be paired off in such a manner that for each pair, the product of their bases is $q^6$ and the product of the corresponding parameters is almost equal to “$a^3$” apart from(with a discrepancy of a small integer power of the base “q”).

Under the replacement $k\to n-k$ on the summation index, these two series are reversal of each other:

$$\begin{array}{cc}\hfill {\Lambda }_n(a,c)=& -q^{7n-1}{a}^3{\lambda }_n(q^{-2n}/a,q^{-n}/c)\frac{{\left[qa,a/q;q\right]}_{2n}{\left[a^{2}c,a^3/c;q^5\right]}_n}{{\left(q{a}^2;q\right)}_{4n}{\left[c,a/c;q\right]}_n},\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill {\lambda }_n(a,c)=& -q^{7n+1}{a}^3{\Lambda }_n(q^{-2n}/a,q^{-n}/c)\frac{{\left(a^2;q\right)}_{4n}{\left[qc,qa/c;q\right]}_n}{{\left[a,q^{2}a;q\right]}_{2n}{\left[q^5{a}^{2}c,q^5{a}^3/c;q^5\right]}_n}.\hfill \end{array}$$

(4)

Observe that for $n\in {\mathbb{N}}_0$, the ${\Lambda }_n$-series terminates above by the numerator parameters and below by the denominator parameters in the following eight different manners:

$$\begin{array}{l}•a^{2}c=q^{-5n}\mathrm{with}a=-1,-q,q^{\frac{1}{2}},q^{\frac{3}{2}};\\ •a/c=q\mathrm{with}a=\pm q^{\frac{-1-5n}{2}},q^{-1-2n},q^{-2-2n}.\end{array}$$

By means of the modified Abel lemma on summation by parts, we shall show, in the next section, a reciprocal formula (see Lemma 1) between the ${\Lambda }_n$-series and the following “twisted cubic series” introduced recently by the author [28]:

$${\Omega }_n(a,c)=\sum _{k=0}^n\left(1-q^{4k}a\right)\frac{{(a/c^2;q)}_{2k}}{{(q{c}^2;q)}_{2k}}\frac{{[c,c,qc;;q]}_k}{{(qa/c;q)}_{3k}}\frac{{(qac;q^5)}_k}{{(1/c;q^{-1})}_k}{q}^k.$$

(5)

When the ${\Lambda }_n$-series terminates above by $a^{2}c=q^{-5n}$, we shall derive four summation formulae (see Theorems 1 and 2), respectively, terminated below by $a=-1,-q$ and $a=q^{1/2},q^{1/2}$. When the ${\Lambda }_n$-series terminates below by $c=a/q$, we find a nonterminating series identity (see Theorem 3) and establish five summation formulae for terminating series (see Theorems 4–6). Finally in Section 3, the corresponding identities for the reversal of ${\lambda }_n$-series will be deduced.

In order to reduce lengthy expressions, we shall utilize the following notations. As usual, the logical function $\chi$ is defined by $\chi \left(\mathrm{true}\right)=1$ and $\chi \left(\mathrm{false}\right)=0$. For a real number x, the greatest integer not exceeding x will be denoted by $\lfloor x\rfloor$. When m is a natural number, $i{\equiv }_{m}j$ signifies that “i is congruent to j modulo m”. To ensure the accuracy, we have checked numerically, throughout the paper, all the displayed equations and identities by appropriately devised Mathematica commands.

2. Identities for the ${\Lambda }_n$-Series

The approach adopted in this paper will be the modified Abel’s lemma on summation by parts (cf. Chu et al. [16,24,29,30,31]). In order to make the paper self–contained, we record it as follows. For an arbitrary complex sequence $\left\{{\tau }_k\right\}$, the backward and forward difference operators ∇ and $\Delta ·$ are defined, respectively, by

$$\begin{array}{c}\hfill \nabla {\tau }_k={\tau }_k-{\tau }_{k-1}\mathrm{and}\Delta ·{\tau }_k={\tau }_k-{\tau }_{k+1},\end{array}$$

where $\Delta ·$ is designed for convenience in the present paper, which differs from the usual operator $\Delta$ only in the minus sign. Then, Abel’s lemma on summation by parts can be modified in the following way:

$$\sum _{k=0}^n{V}_k\nabla U_k=\left\{U_n{V}_{n+1}-U_{-1}{V}_0\right\}+\sum _{k=0}^n{U}_k\Delta ·V_k.$$

(6)

In fact, it is almost routine to check the following expression:

$$\sum _{k=0}^n{V}_k\nabla U_k=\sum _{k=0}^n{V}_k\left\{U_k-U_{k-1}\right\}=\sum _{k=0}^n{U}_k{V}_k-\sum _{k=0}^n{U}_{k-1}{V}_k.$$

Replacing k with $k+1$ for the last sum, we can reformulate the equation as follows

$$\begin{array}{ccc}\hfill \sum _{k=0}^n{V}_k\nabla U_k& =& U_n{V}_{n+1}-U_{-1}{V}_0+\sum _{k=0}^n{U}_k\left\{V_k-V_{k+1}\right\}\hfill \\ & =& U_n{V}_{n+1}-U_{-1}{V}_0+\sum _{k=0}^n{U}_k\Delta ·V_k\hfill \end{array}$$

which is exactly the equality highlighted in the modified Abel lemma.

2.1. Transformation from ${\Lambda }_{\mathit{n}}$ to ${\Omega }_{\mathit{n}}$

For the two sequences $\{{\mathtt{U}}_k,{\mathtt{V}}_k\}$ given by

$$\begin{array}{cc}\hfill {\mathtt{U}}_k:=& \frac{{(qa;q^2)}_k{(q^{2}a;q^2)}_k}{{(q^3{a}^2;q^4)}_k{(q^4{a}^2;q^4)}_k}\frac{{(q^2{a}^2/c;q^3)}_k}{{(q^{3}ac;q^3)}_k}\frac{{(q^5{a}^{2}c;q^5)}_k}{{(a/c;q)}_k},\hfill \\ \hfill {\mathtt{V}}_k:=& \frac{{(a;q^2)}_k{(qa;q^2)}_k}{{(q{a}^2;q^4)}_k{(q^2{a}^2;q^4)}_k}\frac{{(q^{3}ac;q^3)}_k}{{(q^{-1}{a}^2/c;q^3)}_k}\frac{{(a^3/c;q^5)}_k}{{(q^{2}c;q)}_k};\hfill \end{array}$$

it is not difficult to compute their differences

$$\begin{array}{cc}\hfill \nabla {\mathtt{U}}_k:=& \frac{{(a;q^2)}_k{(a/q;q^2)}_k}{{(q^3{a}^2;q^4)}_k{(q^4{a}^2;q^4)}_k}\frac{{(q^{-1}{a}^2/c;q^3)}_k}{{(q^{3}ac;q^3)}_k}\frac{{(a^{2}c;q^5)}_k}{{(a/c;q)}_k}\hfill \\ \hfill \times & \left(\frac{q^k}{-c}\right){\left[\begin{array}{c}{q}^{6k-1}{a}^3,q^{2k}a,q^{k}c,q^{k+1}c\\ a,q/a,a^{2}c,q^{-1}{a}^2/c\end{array}|q\right]}_1,\hfill \\ \hfill \Delta ·{\mathtt{V}}_k:=& \frac{{(a;q^2)}_k{(qa;q^2)}_k}{{(q^5{a}^2;q^4)}_k{(q^6{a}^2;q^4)}_k}\frac{{(q^{3}ac;q^3)}_k}{{(q^2{a}^2/c;q^3)}_k}\frac{{(a^3/c;q^5)}_k}{{(q^{3}c;q)}_k}\hfill \\ \hfill \times & q^k{\left[\begin{array}{c}{q}^{2+6k}{a}^3,q^{1+2k}a,q^{k-1}a/c,q^{k-2}a/c\\ q{a}^2,q^2{a}^2,q^{-1}{a}^2/c,q^{-2}/c\end{array}|q\right]}_1;\hfill \end{array}$$

and the two boundary values

$$\begin{array}{cc}\hfill {\mathtt{U}}_n{\mathtt{V}}_{n+1}=& \frac{{[q^5{a}^{2}c,q^5{a}^3/c;q^5]}_n}{{[q^{3}c,a/c;q]}_n}\frac{{[qa,q^{2}a;q]}_{2n}}{{(q^3{a}^2;q)}_{4n}}\hfill \\ \hfill \times & {\left[\begin{array}{c}{q}^{3+3n}ac,a,qa,a^3/c\\ q{a}^2,q^2{a}^2,q^{2}c,q^{-1}{a}^2/c\end{array}|q\right]}_1,\hfill \\ \hfill {\mathtt{U}}_{-1}{\mathtt{V}}_0=& {\left[\begin{array}{c}ac,a^2,a^2/q,q^{-1}a/c\\ a,a/q,a^{2}c,q^{-1}{a}^2/c\end{array}|q\right]}_1.\hfill \end{array}$$

Observe that the ${\Lambda }_n$-sum can be expressed as

$$\begin{array}{cc}\hfill {\Lambda }_n(a,c)=& \sum _{k=0}^n\left(1-q^{6k-1}{a}^3\right)\frac{{[a^{2}c,a^3/c;q^5]}_k}{{[c,a/c;q]}_k}\frac{{[qa,a/q;q]}_{2k}}{{(q{a}^2;q)}_{4k}}{q}^k\hfill \\ \hfill =& {\left[\begin{array}{c}q/a,a^{2}c,a^2/cq\\ qc,1/c\end{array}|q\right]}_1\times \sum _{k=0}^n{\mathtt{V}}_k\nabla {\mathtt{U}}_k.\hfill \end{array}$$

Then by making use of the modified Abel lemma on summation parts, we can manipulate it as follows:

$$\begin{array}{cc}\hfill {\Lambda }_n(a,c)=& {\left[\begin{array}{c}q/a,a^{2}c,a^2/qc\\ qc,1/c\end{array}|q\right]}_1\hfill \\ \hfill \times & \left\{{\mathtt{U}}_n{\mathtt{V}}_{n+1}-{\mathtt{U}}_{-1}{\mathtt{V}}_0+\sum _{k=0}^n{\mathtt{U}}_k\Delta ·{\mathtt{V}}_k\right\}.\hfill \end{array}$$

After substituting the related terms into the last expression, we can restate it as the recurrence relation below:

$$\begin{array}{cc}\hfill {\Lambda }_n(a,c)=& {\Lambda }_n(qa,q^{3}c){\left[\begin{array}{c}qa,a,a/q,a^{2}c\\ a^{-1},c,qc,q^{2}c\end{array}|q\right]}_1\frac{{(qc/a;q)}_2}{{(q{a}^2;q)}_2}\hfill \\ \hfill +& {\left[\begin{array}{c}ac,a^2,a^2/q,qc/a\\ a,c,qc\end{array}|q\right]}_1+\left(\frac{qc}{a}\right)\frac{{[q^5{a}^{2}c,q^5{a}^3/c;q^5]}_n}{{[q^{3}c,a/c;q]}_n}\hfill \\ & \times {\left[\begin{array}{c}{q}^{3+3n}ac,qa,a,a/q,a^{2}c,a^3/c\\ c,qc,q^{2}c,q{a}^2,q^2{a}^2\end{array}|q\right]}_1\frac{{[qa,q^{2}a;q]}_{2n}}{{(q^3{a}^2;q)}_{4n}}.\hfill \end{array}$$

By iterating the relation m-times and then simplifying the resulting equation, we obtain the following transformation formula.

Lemma 1

(Reciprocal relation). Between the quintic ${\Lambda }_n$-series (1) and the cubic ${\Omega }_m$-series (5), there holds

$$\begin{array}{cc}{\Lambda }_n(a,c)=\hfill & {\Lambda }_n(q^{m}a,q^{3m}c)\frac{{(qc/a;q)}_{2m}{[qa,a,a/q;q]}_m{(a^{2}c;q^5)}_m}{{(q{a}^2;q)}_{2m}{(c;q)}_{3m}{(a^{-1};q^{-1})}_m}\hfill \\ \hfill +& t{\Omega }_{m-1}(ac,a/q){\left[\begin{array}{c}{a}^2,a^2/q,qc/a\\ a,c,qc\end{array}|q\right]}_1+{\Omega }_{m-1}(q^{3+3n}ac,q^{1+2n}a)\hfill \\ & \times \left(\frac{qc}{a}\right)\frac{{(qa;q)}_{2n+1}{(a/q;q)}_{2n+2}{\left[a^{2}c,a^3/c;q^5\right]}_{n+1}}{{(a/c;q)}_n{(c;q)}_{n+3}{(q{a}^2;q)}_{4n+2}}.\end{array}$$

2.2. Case $\boxed{{\mathit{a}}^{\mathbf{2}}\mathit{c}={\mathit{q}}^{-\mathbf{5}\mathit{n}}}$

When $m=n+1$ and $a^{2}c=q^{-5n}$ in Lemma 1, the first term and the third term on the right vanish. We find, in this case, the following simplified transformation formula.

Proposition 1

($a^{2}c=q^{-5n}$ with $n\in {\mathbb{N}}_0$).

$$\begin{array}{c}\hfill {\Lambda }_n(a,c)={\Omega }_n(ac,a/q){\left[\begin{array}{c}{a}^2,a^2/q,qc/a\\ a,c,qc\end{array}|q\right]}_1.\end{array}$$

Letting $a=-q$ and $a=q^{\frac{3}{2}}$ in Proposition 1, we obtain two reduced relations

$$\begin{array}{ll}\hfill {\Lambda }_n(-q,q^{-2-5n})=& {\Omega }_n(-q^{-1-5n},-1){\left[\begin{array}{c}q,q^2,-q^{-2-5n}\\ -q,q^{-1-5n},q^{-2-5n}\end{array}|q\right]}_1,\hfill \\ \hfill {\Lambda }_n(q^{\frac{3}{2}},q^{-3-5n})=& {\Omega }_n(q^{-\frac{3}{2}-5n},q^{\frac{1}{2}}){\left[\begin{array}{c}{q}^2,-q^{\frac{3}{2}},q^{-\frac{7}{2}-5n}\\ q^{-2-5n},q^{-3-5n}\end{array}|q\right]}_1.\hfill \end{array}$$

By combining them with the two summation formulae for the twisted cubic series (Chu [28], Corollary 3)

$$\begin{array}{cc}& {\Omega }_n(-q^{-1-5n},-1)=q^{-1-5n}\frac{{(-q;q)}_{5n+1}{(q^5;q^5)}_n}{{(q;q)}_{5n}{(-q^5;q^5)}_n},\hfill \\ & =\sum _{k=0}^n\left(1+q^{4k-5n-1}\right)\frac{{(q^{-5n};q^5)}_k}{{(q;q)}_{2k}}\frac{{(-q^{-1-5n};q)}_{2k}}{{(q^{-5n};q)}_{3k}}\frac{{\left[-1,-1,-q;q\right]}_k}{{(-1;q^{-1})}_k}{q}^k,\hfill \end{array}$$

(7)

$$\begin{array}{cc}& {\Omega }_n(q^{-\frac{3}{2}-5n},q^{\frac{1}{2}})=\frac{-1}{q^{5n+3/2}}\frac{{(q^5;q^5)}_n{(q^{3/2};q)}_{5n+1}}{{(q^{5/2};q^5)}_n{(q;q)}_{5n+1}},\hfill \\ & =\sum _{k=0}^n\left(1-q^{4k-5n-\frac{3}{2}}\right)\frac{{(q^{-5n};q^5)}_k}{{(q;q)}_{2k+1}}\frac{{(q^{-5n-\frac{5}{2}};q)}_{2k}}{{(q^{-5n-1};q)}_{3k}}\frac{{\left[q^{\frac{1}{2}},q^{\frac{1}{2}},q^{\frac{3}{2}};q\right]}_k}{{(q^{-\frac{1}{2}};q^{-1})}_k}{q}^k.\hfill \end{array}$$

(8)

we find the following explicit formulae.

Theorem 1

((Chu [27], Equation 5.4d: $\epsilon =2$) for the second identity).

$$\begin{array}{cc}\hfill \left(a\right){\Lambda }_n(-q,q^{-2-5n})& =\sum _{k=0}^n\left(1+q^{6k+2}\right)\frac{{[q^{-5n},-q^{5n+5};q^5]}_k}{{[q^{-2-5n},-q^{3+5n};q]}_k}\frac{{[-q^2,-1;q]}_{2k}}{{(q;q)}_{4k+2}}{q}^k\hfill \\ & =\frac{{(q^5;q^5)}_n{(-q^2;q)}_{5n+1}}{{(-q^5;q^5)}_n{(q;q)}_{5n+2}},\hfill \\ \hfill \left(b\right){\Lambda }_n(q^{\frac{3}{2}},q^{-3-5n})& =\sum _{k=0}^n\left(1-q^{6k+\frac{7}{2}}\right)\frac{{[q^{-5n},q^{\frac{15}{2}+5n};q^5]}_k}{{[q^{-3-5n},q^{\frac{9}{2}+5n};q]}_k}\frac{{[q^{\frac{1}{2}},q^{\frac{5}{2}};q]}_{2k}}{{(q;q)}_{4k+3}}{q}^k\hfill \\ & =\frac{{(q^5;q^5)}_n{(q^{\frac{5}{2}};q)}_{5n+2}}{{(q^{\frac{5}{2}};q^5)}_{n+1}{(q;q)}_{5n+3}}.\hfill \end{array}$$

When $a\to -1$ and $a\to q^{\frac{1}{2}}$, the factorial ${(a^2/q;q)}_{2k}$ becomes a zero denominator in ${\Omega }_n(ac,a/q)$. Pulling out it by shifting forward the summation index $k\to k+1$, we can evaluate the following limits:

$$\begin{array}{cc}\hfill {\Lambda }_n(-1,q^{-5n})=& \underset{a\to -1}{lim}{\Omega }_n(q^{-5n}/a,a/q){\left[\begin{array}{c}{a}^2,a^2/q,q^{1-5n}/a^3\\ a,q^{-5n}/a^2,q^{1-5n}/a^2\end{array}|q\right]}_1\hfill \\ \hfill =& \underset{a\to -1}{lim}{\Omega }_{n-1}(q^{4-5n}/a,a)\frac{a^2(1-q/a)(1-q^{-5n}){(q^{1-5n}/a^3;q)}_3}{q{(q^{-5n}/a^2;q)}_5}\hfill \\ \hfill =& {\Omega }_{n-1}(-q^{4-5n},-1)\times \frac{(1+q^{-1}){(-q^{1-5n};q)}_3}{{(q^{1-5n};q)}_4};\hfill \\ \hfill {\Lambda }_n(q^{\frac{1}{2}},q^{-1-5n})=& \underset{a\to q^{\frac{1}{2}}}{lim}{\Omega }_n(q^{-5n}/a,a/q){\left[\begin{array}{c}{a}^2,a^2/q,q^{1-5n}/a^3\\ a,q^{-5n}/a^2,q^{1-5n}/a^2\end{array}|q\right]}_1\hfill \\ \hfill =& \underset{a\to q^{\frac{1}{2}}}{lim}{\Omega }_{n-1}(q^{4-5n}/a,a)\frac{a^2(1-q/a)(1-q^{-5n}){(q^{1-5n}/a^3;q)}_3}{q{(q^{-5n}/a^2;q)}_5}\hfill \\ \hfill =& {\Omega }_{n-1}(q^{\frac{7}{2}-5n},q^{\frac{1}{2}})\times \frac{(1-q^{\frac{1}{2}}){(q^{-\frac{1}{2}-5n};q)}_3}{(1-q^{-1-5n}){(q^{1-5n};q)}_3}.\hfill \end{array}$$

By making use of (7) and (8) under the replacement $n\to n-1$, we derive two further summation formulae.

Theorem 2

((Chu [27], Equation 5.4d: $\epsilon =1$) for the second identity).

$$\begin{array}{cccc}\hfill \left(a\right)& {\Lambda }_n(-1,q^{-5n})\hfill & & =\sum _{k=0}^n\left(1+q^{6k-1}\right)\frac{{[q^{-5n},-q^{5n};q^5]}_k}{{[q^{-5n},-q^{5n};q]}_k}\frac{{[-q,-1/q;q]}_{2k}}{{(q;q)}_{4k}}{q}^k,\hfill \\ & & & =\frac{{(q^5;q^5)}_n{(-q;q)}_{5n}}{{(-q^5;q^5)}_n{(q;q)}_{5n}}(1+q^{-1}),\hfill \\ \hfill \left(b\right)& {\Lambda }_n(q^{\frac{1}{2}},q^{-1-5n})\hfill & & =\sum _{k=0}^n\left(1-q^{6k+\frac{1}{2}}\right)\frac{{[q^{-5n},q^{\frac{5}{2}+5n};q^5]}_k}{{[q^{-1-5n},q^{\frac{3}{2}+5n};q]}_k}\frac{{[q^{\frac{3}{2}},q^{-\frac{1}{2}};q]}_{2k}}{{(q;q)}_{4k+1}}{q}^k\hfill \\ & & & =\frac{{(q^5;q^5)}_n{(q^{\frac{1}{2}};q)}_{5n+1}}{{(q^{\frac{5}{2}};q^5)}_n{(q;q)}_{5n+1}}.\hfill \end{array}$$

2.3. Case $\boxed{\mathit{a}/\mathit{c}=\mathit{q}}$

When $m=1+\lfloor \frac{n}{2}\rfloor$ and $c=a/q$ in Lemma 1, the first term and the second term on the right will be annihilated. We obtain, in this case, the following simplified reciprocal relation.

Proposition 2

(Transformation formula). Between the quintic ${\Lambda }_n$-series (3) and the cubic ${\Omega }_m$-series (5), there holds

$${\Lambda }_n(a,q)={\Omega }_{\lfloor \frac{n}{2}\rfloor }(q^{2+3n}{a}^2,q^{1+2n}a)\times \frac{{(qa;q)}_{2n+1}{(a/q;q)}_{2n+2}{\left[q{a}^2,a^3/q;q^5\right]}_{n+1}}{{(q;q)}_n{(a/q;q)}_{n+3}{(q{a}^2;q)}_{4n+2}}.$$

Letting $n\to \infty$ in Proposition 2 and then invoking the Weierstrass M-test on uniformly convergent series (cf. Stromberg ([32], §3.106)), we find the following beautiful infinite series identity.

Theorem 3

(Nonterminating series identity).

$$\begin{array}{cc}\hfill {\Lambda }_{\infty }(a,q)& =\sum _{k=0}^{\infty }\left(1-q^{6k-1}{a}^3\right)\frac{{[q{a}^2,a^3/q;q^5]}_k}{{[q,a/q;q]}_k}\frac{{[qa,a/q;q]}_{2k}}{{(q{a}^2;q)}_{4k}}{q}^k\hfill \\ & ={\left[\begin{array}{c}qa\\ q,q{a}^2\end{array}|q\right]}_{\infty }{\left[q{a}^2,a^3/q;q^5\right]}_{\infty }\sum _{n=0}^{\infty }{q}^{5\left(\genfrac{}{}{0pt}{}{n}{2}\right)+3n}{(-a)}^n.\hfill \end{array}$$

When $a=-q^2$, the last sum can be evaluated in closed form (cf. Bailey ([33], §8.6))

$$\sum _{n=0}^{\infty }{q}^{5\left(\genfrac{}{}{0pt}{}{n+1}{2}\right)}=\frac{{(q^{10};q^{10})}_{\infty }}{{(q^5;q^{10})}_{\infty }}.$$

This leads us to a very strange infinite series identity.

Corollary 1

(Nonterminating series identity).

$$\begin{array}{cc}\hfill {\Lambda }_{\infty }(-q^2,q)& =\sum _{k=0}^{\infty }\left(1+q^{6k+5}\right)\frac{{(q^{10};q^{10})}_k}{{(q^2;q^2)}_k}\frac{{(-q;q)}_{2k}{(-q;q)}_{2k+2}}{{(q;q)}_{4k+4}}{q}^k\hfill \\ & =\frac{{(-q;q)}_{\infty }{(q^{10};q^{10})}_{\infty }^2}{{(q;q)}_{\infty }^2{(q^5;q^{10})}_{\infty }}.\hfill \end{array}$$

Letting $a=q^{\frac{1-5n}{3}}$ in Proposition 2, we deduce immediately the limiting result.

Theorem 4

(Limiting relation).

$${\Lambda }_n(a,q)\to 0asa\to q^{\frac{1-5n}{3}}.$$

When $a\to \pm q^{\frac{-1-5n}{2}}$ in Proposition 2, the factorial quotient on the right contains a zero factor, which cancels all the terms in ${\Omega }_{\lfloor \frac{n}{2}\rfloor }(q^{2+3n}{a}^2,q^{1+2n}a)$, except for the ultimate one. Consequently, we have the identities.

Theorem 5

(Summation formulae for terminating series).

$$\begin{array}{cc}\hfill \left(a\right)& \mathrm{If}n{\equiv }_{2}0,\mathrm{then}\hfill \\ & {\Lambda }_n(\pm q^{\frac{-1-5n}{2}},q)\hfill \\ & =\sum _{k=0}^n\left(1\mp q^{6k-\frac{5+15n}{2}}\right)\frac{{[q^{-5n},\pm q^{-\frac{5+15n}{2}};q^5]}_k}{{[q,\pm q^{-\frac{3+5n}{2}};q]}_k}\frac{{[\pm q^{\frac{1-5n}{2}},\pm q^{-\frac{3+5n}{2}};q]}_{2k}}{{(q^{-5n};q)}_{4k}}{q}^k\hfill \\ & ={(\mp 1)}^{\frac{n+2}{2}}\frac{{(q^5;q^5)}_n}{{(q;q)}_{5n}}{(\pm q^{1/2};q)}_{\frac{5n}{2}}{(\pm q^{5/2};q^5)}_{\frac{2+3n}{2}}{q}^{(5n^2-60n-20)/8}.\hfill \\ \hfill \left(b\right)& \mathrm{If}n{\equiv }_{2}1,\mathrm{then}\hfill \\ & {\Lambda }_n(\pm q^{\frac{-1-5n}{2}},q)\hfill \\ & \sum _{k=0}^n\left(1\mp q^{6k-\frac{5+15n}{2}}\right)\frac{{[q^{-5n},\pm q^{-\frac{5+15n}{2}};q^5]}_k}{{[q,\pm q^{-\frac{3+5n}{2}};q]}_k}\frac{{[\pm q^{\frac{1-5n}{2}},\pm q^{-\frac{3+5n}{2}};q]}_{2k}}{{(q^{-5n};q)}_{4k}}{q}^k\hfill \\ & =-(1\mp 1)\frac{{(q^5;q^5)}_n}{{(q;q)}_{5n}}{(-q;q)}_{\frac{5n-1}{2}}{(-q^5;q^5)}_{\frac{1+3n}{2}}{q}^{(5n^2-60n-17)/8}.\hfill \end{array}$$

The two remaining terminating series ${\Lambda }_n(q^{-1-2n},q)$ and ${\Lambda }_n(q^{-2-2n},q)$ are much harder to evaluate, whose closed expressions are given as follows.

Theorem 6

(Summation formulae for terminating series).

$$\begin{array}{cc}\hfill \left(a\right)& {\Lambda }_n(q^{-1-2n},q)\hfill \\ & =\sum _{k=0}^n\left(1-q^{6k-6n-4}\right)\frac{{[q^{-1-4n},q^{-4-6n};q^5]}_k}{{[q,q^{-2-2n};q]}_k}\frac{{[q^{-2n},q^{-2-2n};q]}_{2k}}{{(q^{-1-4n};q)}_{4k}}{q}^k\hfill \\ & =\chi (n{\equiv }_{5}0,2)\frac{{(q^{-1-4n};q^5)}_{2n+2}}{{(q^{2n+1};q)}_{2n+1}}{q}^{2n^2-3n-3}.\hfill \\ \hfill \left(b\right)& {\Lambda }_n(q^{-2-2n},q)\hfill \\ & =\sum _{k=0}^n\left(1-q^{6k-6n-7}\right)\frac{{[q^{-3-4n},q^{-7-6n};q^5]}_k}{{[q,q^{-3-2n};q]}_k}\frac{{[q^{-1-2n},q^{-3-2n};q]}_{2k}}{{(q^{-3-4n};q)}_{4k}}{q}^k\hfill \\ & =\chi (n{\equiv }_{5}0,1)\frac{{(q^{-3-4n};q^5)}_{2n+3}}{{(q^{2n+2};q)}_{2n+2}}{q}^{2n^2-n-4}.\hfill \end{array}$$

Proof. 

The difficulty to prove the theorem lies in the fact that there are zero factors appearing in both numerators and denominators of the cubic series ${\Omega }_{\lfloor \frac{n}{2}\rfloor }$ in Proposition 2. In order to overcome this obstacle, we first reformulate, under the replacement $a=q^{-2-2n}x$, the equality in Proposition 2 as

$${\Lambda }_n(q^{-2-2n}x,q)=P_n\left(x\right)\times Q_n\left(x\right)$$

where $P_n\left(x\right)$ and $Q_n\left(x\right)={\Omega }_{\lfloor \frac{n}{2}\rfloor }(q^{-2-n}{x}^2,x)$ are given explicitly by

$$\begin{array}{cc}\hfill P_n\left(x\right)=& \frac{{(q^{-1-2n}x;q)}_{2n+1}{(q^{-3-2n}x;q)}_{2n+2}{\left[q^{-3-4n}{x}^2,q^{-7-6n}{x}^3;q^5\right]}_{n+1}}{{(q;q)}_n{(q^{-3-2n}x;q)}_{n+3}{(q^{-3-4n}{x}^2;q)}_{4n+2}},\hfill \\ \hfill Q_n\left(x\right)=& \sum _{k=0}^{\lfloor \frac{n}{2}\rfloor }\left(1-q^{4k-2-n}{x}^2\right)\frac{{(q^{-n};q)}_{2k}}{{(x^2/q;q)}_{2k}}\frac{{\left[x,x/q,x/q;q\right]}_k{(q^{-2-n}{x}^3;q^5)}_k}{{(q/x;q^{-1})}_k{(q^{-n}x;q)}_{3k}}{q}^k.\hfill \end{array}$$

Then it is not hard to evaluate the limits:

$$\begin{array}{cc}\hfill \underset{x\to q}{lim}{P}_n\left(x\right)=& {(-1)}^n\frac{{\left[q^{1-n},q^{4+n};q^5\right]}_{n+1}}{2(1-q^n){(q^{2n+1};q)}_{2n+1}}{q}^{\frac{n^2-9n-8}{2}},\hfill \\ \hfill \underset{x\to 1}{lim}{P}_n\left(x\right)=& {(-1)}^n\frac{{\left[q^{3-n},q^{7+n};q^5\right]}_{n+1}}{{(q^{2n+2};q)}_{2n+2}}{q}^{\frac{n^2-9n-10}{2}}.\hfill \end{array}$$

When $x\to q$, the only nonzero terms in $Q_n\left(x\right)$ are those with their summation index $k=0$ and $\lfloor \frac{n+2}{3}\rfloor \le k\le \lfloor \frac{n}{2}\rfloor$. Writing the initial term separately and then shifting forward the summation index by $k\to k+\lfloor \frac{n+2}{3}\rfloor$, we have the expression

$$\begin{array}{cc}\hfill \underset{x\to q}{lim}{Q}_n\left(x\right)=& (1-q^{-n})+{\Omega }_{\lfloor \frac{n}{2}\rfloor -\lfloor \frac{n+2}{3}\rfloor }(q^{4\lfloor \frac{n+2}{3}\rfloor -n},q^{\lfloor \frac{n+2}{3}\rfloor })\hfill \\ \hfill \times & q^{\lfloor \frac{n+2}{3}\rfloor }\underset{x\to q}{lim}\frac{{(q^{-n};q)}_{2\lfloor \frac{n+2}{3}\rfloor }}{{(x^2/q;q)}_{2\lfloor \frac{n+2}{3}\rfloor }}\frac{{\left[x,x/q,x/q;q\right]}_{\lfloor \frac{n+2}{3}\rfloor }{(q^{-2-n}{x}^3;q^5)}_{\lfloor \frac{n+2}{3}\rfloor }}{{(q/x;q^{-1})}_{\lfloor \frac{n+2}{3}\rfloor }{(q^{-n}x;q)}_{3\lfloor \frac{n+2}{3}\rfloor }}\hfill \\ \hfill =& (1-q^{-n})+\rho \left(n\right)\times {\Omega }_{\lfloor \frac{n}{2}\rfloor -\lfloor \frac{n+2}{3}\rfloor }(q^{4\lfloor \frac{n+2}{3}\rfloor -n},q^{\lfloor \frac{n+2}{3}\rfloor }),\hfill \end{array}$$

where $\rho \left(n\right)$ is given by

$$\rho \left(n\right)=\frac{(1-q^n){(q^{1-n};q^5)}_{\lfloor \frac{n+2}{3}\rfloor }}{(1-q){(q^{\lfloor \frac{n+5}{3}\rfloor };q)}_{\lfloor \frac{n+2}{3}\rfloor }}\times \left\{\begin{array}{cc}-\frac{1-q}{1-q^{\frac{n}{3}}}{q}^{\frac{n^2-6n}{9}},\hfill & n{\equiv }_{3}0;\hfill \\ -\frac{1-q^{\frac{n-1}{3}}}{1-q^2}{q}^{\frac{n^2-8n+7}{9}},\hfill & n{\equiv }_{3}1;\hfill \\ q^{\frac{n^2-7n+1}{9}},\hfill & n{\equiv }_{3}2.\hfill \end{array}\right.$$

Analogously for $x\to 1$, the nonzero terms in $Q_n\left(x\right)$ are those with $k=0,1$ and $\lfloor \frac{n+3}{3}\rfloor \le k\le \lfloor \frac{n}{2}\rfloor$. Putting the first two terms together and then shifting forward the summation index by $k\to k+\lfloor \frac{n+3}{3}\rfloor$, we have another expression

$$\begin{array}{cc}\hfill \underset{x\to 1}{lim}{Q}_n\left(x\right)=& \frac{1-q^{-2-n}}{2}+{\Omega }_{\lfloor \frac{n}{2}\rfloor -\lfloor \frac{n+3}{3}\rfloor }(q^{2+4\lfloor \frac{n}{3}\rfloor -n},q^{\lfloor \frac{n}{3}\rfloor })\hfill \\ \hfill \times & q^{\lfloor \frac{n+3}{3}\rfloor }\underset{x\to 1}{lim}\frac{{(q^{-n};q)}_{2\lfloor \frac{n+3}{3}\rfloor }}{{(x^2/q;q)}_{2\lfloor \frac{n+3}{3}\rfloor }}\frac{{\left[x,x/q,x/q;q\right]}_{\lfloor \frac{n+3}{3}\rfloor }{(q^{-2-n}{x}^3;q^5)}_{\lfloor \frac{n+3}{3}\rfloor }}{{(q/x;q^{-1})}_{\lfloor \frac{n+3}{3}\rfloor }{(q^{-n}x;q)}_{3\lfloor \frac{n+3}{3}\rfloor }}\hfill \\ \hfill =& \frac{1-q^{-2-n}}{2}+\varrho \left(n\right)\times {\Omega }_{\lfloor \frac{n}{2}\rfloor -\lfloor \frac{n+3}{3}\rfloor }(q^{2+4\lfloor \frac{n}{3}\rfloor -n},q^{\lfloor \frac{n}{3}\rfloor }),\hfill \end{array}$$

where $\varrho \left(n\right)$ is given by

$$\varrho \left(n\right)=\frac{{(q^{-2-n};q^5)}_{\lfloor \frac{n+3}{3}\rfloor }}{2{(q^{\lfloor \frac{n+3}{3}\rfloor };q)}_{\lfloor \frac{n}{3}\rfloor }}\times \left\{\begin{array}{cc}-\frac{(1-q^{\frac{n-3}{3}})q^{\frac{n^2-3n+9}{9}}}{(1-q)(1-q^2)},\hfill & n{\equiv }_{3}0;\hfill \\ \frac{q^{\frac{n^2-2n+1}{9}}}{1-q},\hfill & n{\equiv }_{3}1;\hfill \\ \frac{-q^{\frac{n^2-n-2}{9}}}{1-q^{\frac{n-2}{3}}},\hfill & n{\equiv }_{3}2.\hfill \end{array}\right.$$

Consequently, we have established the following reduction formulae:

$$\begin{array}{cc}\hfill {\Lambda }_n(q^{-1-2n},q)=& \underset{x\to q}{lim}{P}_n\left(x\right)\times Q_n\left(x\right)\hfill \\ \hfill =& {(-1)}^n\frac{{\left[q^{1-n},q^{4+n};q^5\right]}_{n+1}}{2(1-q^n){(q^{2n+1};q)}_{2n+1}}{q}^{\frac{n^2-9n-8}{2}},\hfill \\ \hfill \times & \left\{(1-q^{-n})+\rho \left(n\right){\Omega }_{\lfloor \frac{n}{2}\rfloor -\lfloor \frac{n+2}{3}\rfloor }\left(q^{4\lfloor \frac{n+2}{3}\rfloor -n},q^{\lfloor \frac{n+2}{3}\rfloor }\right)\right\},\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill {\Lambda }_n(q^{-2-2n},q)=& \underset{x\to 1}{lim}{P}_n\left(x\right)\times Q_n\left(x\right)\hfill \\ \hfill =& {(-1)}^n\frac{{\left[q^{3-n},q^{7+n};q^5\right]}_{n+1}}{{(q^{2n+2};q)}_{2n+2}}{q}^{\frac{n^2-9n-10}{2}}\hfill \\ \hfill \times & \left\{\frac{1-q^{-2-n}}{2}+\varrho \left(n\right){\Omega }_{\lfloor \frac{n}{2}\rfloor -\lfloor \frac{n+3}{3}\rfloor }\left(q^{2+4\lfloor \frac{n}{3}\rfloor -n},q^{\lfloor \frac{n}{3}\rfloor }\right)\right\}.\hfill \end{array}$$

(10)

Classifying n according to its residues moduolo 6, we can tabulate the afore-displayed two cubic series with their explicit parameters:

$$n$$	$${\Omega }_{\lfloor \frac{n}{2}\rfloor -\lfloor \frac{n+2}{3}\rfloor }\left(q^{4\lfloor \frac{n+2}{3}\rfloor -n},q^{\lfloor \frac{n+2}{3}\rfloor }\right)$$	$${\Omega }_{\lfloor \frac{n}{2}\rfloor -\lfloor \frac{n+3}{3}\rfloor }\left(q^{2+4\lfloor \frac{n}{3}\rfloor -n},q^{\lfloor \frac{n}{3}\rfloor }\right)$$
$$6m$$	$${\Omega }_m(q^{2m},q^{2m})$$	$${\Omega }_{m-1}(q^{2+2m},q^{2m})$$
$$6m+1$$	$${\Omega }_{m-1}(q^{3+2m},q^{1+2m})$$	$${\Omega }_{m-1}(q^{1+2m},q^{2m})$$
$$6m+2$$	$${\Omega }_m(q^{2+2m},q^{1+2m})$$	$${\Omega }_m(q^{2m},q^{2m})$$
$$6m+3$$	$${\Omega }_m(q^{1+2m},q^{1+2m})$$	$${\Omega }_{m-1}(q^{3+2m},q^{1+2m})$$
$$6m+4$$	$${\Omega }_m(q^{4+2m},q^{2+2m})$$	$${\Omega }_m(q^{2+2m},q^{1+2m})$$
$$6m+5$$	$${\Omega }_m(q^{3+2m},q^{2+2m})$$	$${\Omega }_m(q^{1+2m},q^{1+2m})$$

All these cubic series can be evaluated in closed forms by the three known formulae due to the author ([28], Theorem 10):

$$\begin{array}{cc}\hfill {\Omega }_{\infty }(c,c)& =\frac{{(c;q)}_{\infty }{\left[q^5,q^{3}c,q^2/c,q{c}^2,q^4{c}^3;q^5\right]}_{\infty }}{{\left[q,q{c}^2;q\right]}_{\infty }}\hfill \\ & =\sum _{k=0}^{\infty }\left(1-q^{4k}c\right)\frac{{\left[c,c,qc;q\right]}_k}{{(q;q)}_{3k}}\frac{{(1/c;q)}_{2k}}{{(q{c}^2;q)}_{2k}}\frac{{(q{c}^2;q^5)}_k}{{(1/c;q^{-1})}_k}{q}^k,\hfill \\ \hfill {\Omega }_{\infty }(qc,c)& =\frac{{(qc;q)}_{\infty }{\left[q^5,qc,q^4/c,q^2{c}^2,q^3{c}^3;q^5\right]}_{\infty }}{{\left[q,q{c}^2;q\right]}_{\infty }}\hfill \\ & =\sum _{k=0}^{\infty }\left(1-q^{1+4k}c\right)\frac{{\left[c,c,qc;q\right]}_k}{{(q;q)}_{3k+1}}\frac{{(q/c;q)}_{2k}}{{(q{c}^2;q)}_{2k}}\frac{{(q^2{c}^2;q^5)}_k}{{(1/c;q^{-1})}_k}{q}^k,\hfill \\ \hfill {\Omega }_{\infty }(q^{2}c,c)& =\frac{{(qc;q)}_{\infty }{\left[q^5,q^{4}c,q^6/c,q^3{c}^2,q^2{c}^3;q^5\right]}_{\infty }}{{\left[q,q{c}^2;q\right]}_{\infty }}\hfill \\ & =\sum _{k=0}^{\infty }\left(1-q^{2+4k}c\right)\frac{{\left[c,c,qc;q\right]}_k}{{(q;q)}_{3k+2}}\frac{{(q^2/c;q)}_{2k}}{{(q{c}^2;q)}_{2k}}\frac{{(q^3{c}^2;q^5)}_k}{{(1/c;q^{-1})}_k}{q}^k.\hfill \end{array}$$

After long and tedious computations, we obtain the simplified expressions

$$\begin{array}{cc}\hfill \rho \left(n\right)& \times {\Omega }_{\lfloor \frac{n}{2}\rfloor -\lfloor \frac{n+2}{3}\rfloor }(q^{4\lfloor \frac{n+2}{3}\rfloor -n},q^{\lfloor \frac{n+2}{3}\rfloor })=(1-q^{-n})\times \left\{\begin{array}{cc}+1,\hfill & n{\equiv }_{5}0,2;\hfill \\ 0,\hfill & n{\equiv }_{5}1;\hfill \\ -1,\hfill & n{\equiv }_{5}3,4.\hfill \end{array}\right.\hfill \\ \hfill \varrho \left(n\right)& \times {\Omega }_{\lfloor \frac{n}{2}\rfloor -\lfloor \frac{n+3}{3}\rfloor }(q^{2+4\lfloor \frac{n}{3}\rfloor -n},q^{\lfloor \frac{n}{3}\rfloor }=\frac{1-q^{-2-n}}{2})\times \left\{\begin{array}{cc}+1,\hfill & n{\equiv }_{5}0,1;\hfill \\ 0,\hfill & n{\equiv }_{5}3;\hfill \\ -1,\hfill & n{\equiv }_{5}2,4.\hfill \end{array}\right.\hfill \end{array}$$

Substituting them into (9) and (10) and then making some routine simplifications, we confirm the two summation formulae in Theorem 6. □

3. Closed Formulae for the ${\mathbf{\lambda }}_n$-Series

For the terminating series identities for ${\Lambda }_n$ established in the last section, their counterparts for the series ${\lambda }_m$ can analogously be deduced in view of the reciprocal relations (3) and (4). We illustrate the following example to show how this approach works. For $m\in {\mathbb{N}}_0$ with $m{\not\equiv }_{3}2$, we have, according to (4), the transformation

$$\begin{array}{cc}\hfill {\lambda }_m\left(q^{\frac{-1-m}{3}},q^{-1-m}\right)=& (-q^{6m})\times \underset{x\to 1}{lim}{\Lambda }_m\left(q^{\frac{1-5m}{3}}x,q\right)\hfill \\ \hfill \times & \frac{{(q^{\frac{-2-2m}{3}};q)}_{4m}{\left[q^{-m},q^{\frac{5+2m}{3}};q\right]}_m}{{\left[q^{\frac{-1-m}{3}},q^{\frac{5-m}{3}};q\right]}_{2m}{\left[q^5,q^{\frac{10-5m}{3}};q^5\right]}_m}.\hfill \end{array}$$

Because of $m{\not\equiv }_{3}2$, the above factorial quotient does not have zero factors in its denominator. In view of Theorem 4, we establish the following summation formula.

Corollary 2

(Reversal series from Theorem 4: $m\in {\mathbb{N}}_0$ with $m{\not\equiv }_{3}2$).

$${\lambda }_m\left(q^{\frac{-1-m}{3}},q^{-1-m}\right)=\sum _{k=0}^m\left(1-q^{6k-m}\right)\frac{{[q^{-m},q^{\frac{5+2m}{3}};q]}_k}{{[q^5,q^{\frac{10-5m}{3}};q^5]}_k}\frac{{(q^{-\frac{2+2m}{3}};q)}_{4k}}{{[q^{\frac{5-m}{3}},q^{-\frac{1+m}{3}};q]}_{2k}}{q}^k=0.$$

By examining the reversal series for those given in Theorems 1, 2, 5 and 6, we can further deduce the following summation formulae.

Corollary 3

(Reversal series from Theorem 1).

$$\begin{array}{cc}\hfill \left(a\right)& {\lambda }_m\left(-q^{-1-2m},q^{2+4m}\right)\hfill \\ & =\sum _{k=0}^m\left(1+q^{6k-6m-2}\right)\frac{{[q^{3+4m},-q^{-2-6m};q]}_k}{{[q^5,-q^{-10m};q^5]}_k}\frac{{(q^{-2-4m};q)}_{4k}}{{[-q^{1-2m},-q^{-1-2m};q]}_{2k}}{q}^k\hfill \\ & =\frac{q^{-2-7m-2m^2}{(-q;q)}_{6m+2}}{{(-q;q)}_{2m+1}{(-1;q)}_{2m}{(-q^5;q^5)}_{2m}}.\hfill \\ \hfill \left(b\right)& {\lambda }_m\left(q^{-\frac{3}{2}-2m},q^{3+4m}\right)\hfill \\ & =\sum _{k=0}^m\left(1-q^{6k-6m-\frac{7}{2}}\right)\frac{{[q^{4+4m},q^{-\frac{7}{2}-6m};q]}_k}{{[q^5,q^{-\frac{5}{2}-10m};q^5]}_k}\frac{{(q^{-3-4m};q)}_{4k}}{{[q^{\frac{1}{2}-2m},q^{-\frac{3}{2}-2m};q]}_{2k}}{q}^k\hfill \\ & =\frac{q^{-4-8m-2m^2}{(q^{\frac{1}{2}};q)}_{6m+4}}{{(q^{\frac{3}{2}};q)}_{2m+1}{(q^{-\frac{1}{2}};q)}_{2m+1}{(q^{\frac{5}{2}};q^5)}_{2m+1}}.\hfill \end{array}$$

Corollary 4

(Reversal series from Theorem 2).

$$\begin{array}{cc}\hfill \left(a\right)& {\lambda }_m\left(-q^{-2m},q^{4m}\right)\hfill \\ & =\sum _{k=0}^m\left(1+q^{6k-6m+1}\right)\frac{{[q^{1+4m},-q^{1-6m};q]}_k}{{[q^5,-q^{5-10m};q^5]}_k}\frac{{(q^{-4m};q)}_{4k}}{{[-q^{2-2m},-q^{-2m};q]}_{2k}}{q}^k\hfill \\ & =\frac{q^{1-5m-2m^2}{(-q;q)}_{6m-1}}{{(-q;q)}_{2m}{(-1;q)}_{2m-1}{(-q^5;q^5)}_{2m-1}}.\hfill \\ \hfill \left(b\right)& {\lambda }_m\left(q^{-\frac{1}{2}-2m},q^{1+4m}\right)\hfill \\ & =\sum _{k=0}^m\left(1-q^{6k-6m-\frac{1}{2}}\right)\frac{{[q^{2+4m},q^{-\frac{1}{2}-6m};q]}_k}{{[q^5,q^{\frac{5}{2}-10m};q^5]}_k}\frac{{(q^{-1-4m};q)}_{4k}}{{[q^{\frac{3}{2}-2m},q^{-\frac{1}{2}-2m};q]}_{2k}}{q}^k\hfill \\ & =\frac{q^{-6m-2m^2}{(q^{\frac{1}{2}};q)}_{6m+1}}{{(q^{\frac{1}{2}};q)}_{2m+1}{(q^{\frac{1}{2}};q)}_{2m-1}{(q^{\frac{5}{2}};q^5)}_{2m}}.\hfill \end{array}$$

Corollary 5

(Reversal series from Theorem 5).

$$\begin{array}{cc}\hfill \left(a\right)& Ifm{\equiv }_{2}0,then\hfill \\ & {\lambda }_m\left(\pm q^{\frac{1+m}{2}},q^{-1-m}\right)\hfill \\ & =\sum _{k=0}^m\left(1\mp q^{6k+\frac{5+3m}{2}}\right)\frac{{[q^{-m},\pm q^{\frac{5+3m}{2}};q]}_k}{{[q^5,\pm q^{\frac{15+5m}{2}};q^5]}_k}\frac{{(q^{1+m};q)}_{4k}}{{[\pm q^{\frac{1+m}{2}},\pm q^{\frac{5+m}{2}};q]}_{2k}}{q}^k\hfill \\ & ={(\mp 1)}^{\frac{m}{2}}\frac{{(\pm q^{\frac{1}{2}};q)}_{\frac{m}{2}}{(\pm q^{\frac{5}{2}};q^5)}_{\frac{m+2}{2}}}{{(\pm q^{\frac{5+m}{2}};q)}_m}{q}^{\frac{m^2+4m}{8}}.\hfill \\ \hfill \left(b\right)& Ifm{\equiv }_{2}1,then\hfill \\ & {\lambda }_m\left(\pm q^{\frac{1+m}{2}},q^{-1-m}\right)\hfill \\ & =\sum _{k=0}^m\left(1\mp q^{6k+\frac{5+3m}{2}}\right)\frac{{[q^{-m},\pm q^{\frac{5+3m}{2}};q]}_k}{{[q^5,\pm q^{\frac{15+5m}{2}};q^5]}_k}\frac{{(q^{1+m};q)}_{4k}}{{[\pm q^{\frac{1+m}{2}},\pm q^{\frac{5+m}{2}};q]}_{2k}}{q}^k\hfill \\ & =(1\mp 1)\frac{{(-q;q)}_{\frac{m-1}{2}}{(-q^5;q^5)}_{\frac{m+1}{2}}}{{(-q^{\frac{5+m}{2}};q)}_m}{q}^{\frac{m^2+4m+3}{8}}.\hfill \end{array}$$

Corollary 6

(Reversal series from Theorem 6).

$$\begin{array}{cc}\hfill \left(a\right)& Ifm{\not\equiv }_{5}1,then\hfill \\ & {\lambda }_m\left(q,q^{-1-m}\right)=\sum _{k=0}^m\left(1-q^{6k+4}\right)\frac{{[q^{-m},q^{3+m};q]}_k}{{[q^{6-m},q^{9+m};q^5]}_k}\frac{{(q^2;q)}_{4k}}{{[q,q^3;q]}_{2k}}{q}^k\hfill \\ & \hfill =q^m\frac{(1-q^2)(1-q^{1-m})(1-q^{4+m})}{(1-q^{1+m})(1-q^{2+m})}\chi (m{\equiv }_{5}0,2).\\ \hfill \left(b\right)& Ifm{\not\equiv }_{5}3,then\hfill \\ & {\lambda }_m\left(q^2,q^{-1-m}\right)=\sum _{k=0}^m\left(1-q^{6k+7}\right)\frac{{[q^{-m},q^{4+m};q]}_k}{{[q^{8-m},q^{12+m};q^5]}_k}\frac{{(q^4;q)}_{4k}}{{[q^2,q^4;q]}_{2k}}{q}^k\hfill \\ & \hfill =q^m\frac{(1-q)(1-q^{3-m})(1-q^{7+m})}{(1-q^{1+m})(1-q^{3+m})}\chi (m{\equiv }_{5}0,1).\end{array}$$