Note on q-Series Identities via the Telescoping Method

Abstract

Several q-series identities were obtained using the telescoping method. These identities were then used to derive double series expressions for well-known zeta-value constants.

1. Introduction

The telescoping method is a simple and elegant way to compute infinite series and sums. However, creating telescoping sums is somewhat more skillful and may be very complex more than one’s expect. Interested readers can refer to [1] for detailed information on creating telescoping sums and some applications. In general, by q-series, we mean expressions with summands that are of the type ${(a;q)}_k$. In this short note, we will pick some simple summands to produce some general q-series identities via the telescoping method.

In the following, we introduce some common notations and terminologies. Throughout this paper, q is fixed with $0<q<1$. The q-integer of an integer k is defined by ${\left[k\right]}_q={\displaystyle \frac{1-q^k}{1-q}}$. The q-shifted factorial ${(a;q)}_{\infty }$ is defined as an infinity product as follows:

$${(a;q)}_{\infty }=\prod _{i=0}^{\infty }(1-a{q}^i).$$

Here, we define ${(a;q)}_k={\displaystyle \frac{{(a;q)}_{\infty }}{{(a{q}^k;q)}_{\infty }}}={\prod }_{i=0}^{k-1}(1-a{q}^i)$, and

$${(a,b,\dots ,c;q)}_k={(a;q)}_k{(b;q)}_k\cdots {(c;q)}_k.$$

And the q-analogue binomial coefficient is defined as

$${\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]}_q={\displaystyle \frac{{(q;q)}_{\alpha }}{{(q;q)}_{\beta }{(q;q)}_{\alpha -\beta }}}={\displaystyle \frac{{(q^{\alpha -\beta +1};q)}_{\beta }}{{(q;q)}_{\beta }}}.$$

In this short note, we also define the operator of minus forward difference ${\Delta }_j$. We mean that when a function of j, say $F\left(j\right)$, is applied to the operator ${\Delta }_j$,

$${\Delta }_{j}F\left(j\right)=F\left(j\right)-F(j+1).$$

The well-known Riemann zeta function $\zeta \left(s\right)$ is defined by the infinite series

$$\zeta \left(s\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^s}}$$

for any complex number s with real part greater than 1. In [2], Chu obtained the double series expressions for $\zeta \left(2\right),\zeta \left(3\right)$ and the Catalan constant G. Actually, Chu proved that

$$\zeta \left(2\right)=\sum _{i,j=1}^{\infty }{\displaystyle \frac{(i-1)!(j-1)!}{(i+j)!}},\zeta \left(3\right)={\displaystyle \frac{1}{3}}\sum _{i,j=1}^{\infty }{\displaystyle \frac{(i-1)!(j-1)!}{(i+j)!}}{H}_{i+j},$$

where $H_n$ is the harmonic number given by $H_n={\sum }_{k=1}^n{\displaystyle \frac{1}{k}}$, and the double series expression for Catalan constant G:

$$G:=\sum _{n=1}^{\infty }{\displaystyle \frac{{(-1)}^{n-1}}{{(2n-1)}^2}}={\displaystyle \frac{1}{8}}\sum _{i,j=1}^{\infty }{\displaystyle \frac{(i-1)!(j-1)!}{{\left(\frac{1}{2}\right)}_{i+j}}},$$

with the rising factorial ${\left(x\right)}_n$ defined by ${\left(x\right)}_n=x(x+1)\cdots (x+n-1)$. Indeed, Chu [2] also proved some more general formulas for Riemann zeta function at positive integer values and for Catalan’s constant via the telescoping method. A little later, Chen [3], by using the telescoping method, obtained two general double-series q-analogue formulas as below

$$\begin{array}{c}\hfill \sum _{i=1}^m\sum _{j=1}^k{\displaystyle \frac{f_q\left(i\right)q^j{\left[ni\right]}_q}{{\left[j\right]}_q{\left[\begin{array}{c}ni+j\\ j\end{array}\right]}_q}}=\sum _{i=1}^m{f}_q\left(i\right)\left(1-{\displaystyle \frac{{(q;q)}_k}{{(q^{ni+1};q)}_k}}\right),\end{array}$$

(1)

and

$$\begin{array}{cc}\hfill & \sum _{i=1}^m\sum _{j=1}^k{\displaystyle \frac{{\left[\begin{array}{c}ni+j\\ j\end{array}\right]}_{q^2}}{{\left[\begin{array}{c}2ni+2j\\ 2j\end{array}\right]}_q}}·{\displaystyle \frac{{\left[2ni\right]}_q{f}_q\left(i\right)q^{2j-1}}{{[2j-1]}_q}}=\sum _{i=1}^m{f}_q\left(i\right)\left(1-{\displaystyle \frac{{(q;q^2)}_k}{{(q^{2ni+1};q^2)}_k}}\right).\hfill \end{array}$$

(2)

Here, $f_q\left(z\right)$ is a q-analogue function of z, and $m,n,k$ are any positive integers. Chen [3] also used the above formulas to obtain many unusual double-series expressions of well-known constants. For example,

$$\begin{array}{cc}\hfill & G=\sum _{i,j=1}^{\infty }{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{\lambda i+j}{j}\right)}{\left(\genfrac{}{}{0pt}{}{2\lambda i+2j}{2j}\right)}}{\displaystyle \frac{2\lambda i{(-1)}^{i-1}}{{(2i-1)}^2(2j-1)}}(\lambda \ge 1),\hfill \\ \hfill & 4log\left(2\right)-\zeta \left(2\right)=\sum _{i,j=1}^{\infty }{\displaystyle \frac{4(2i-2)!(2j-2)!(i+j-1)!}{i!(j-1)!(2i+2j-1)!}},\hfill \\ \hfill & {\displaystyle \frac{\sqrt{3}\pi }{18}}=\sum _{i,j=1}^{\infty }{\displaystyle \frac{(i+j)!i!\left(2j\right)!}{(2i+2j)!j!(2j-1)}}.\hfill \end{array}$$

So, in view of (1) and (2), it is natural to ask, for any integer $t\ge 1$, what is the double series expression of the sum?

$$\sum _{i=1}^m{f}_q\left(i\right)\left(1-{\displaystyle \frac{{(q;q^t)}_k}{{(q^{tni+1};q^t)}_k}}\right).$$

In this paper, we will give an answer from a more general q-series identity.

Theorem 1.

Notations are noted as above. Let $f_q\left(z\right)$ be a q-analogue function of z and $m,n,k$ be any positive integers; then, we have

$$\begin{array}{c}\hfill \sum _{i=1}^m\sum _{j=1}^k{\displaystyle \frac{{(aq,bq;q)}_j}{{(cq,dq;q)}_j}}{\displaystyle \frac{(a+b-c-d)q^j+(cd-ab)q^{2j}}{(1-a{q}^j)(1-b{q}^j)}}{f}_q\left(i\right)=\sum _{i=1}^m{f}_q\left(i\right)\left(1-{\displaystyle \frac{{(aq,bq;q)}_k}{{(cq,dq;q)}_k}}\right),\end{array}$$

(3)

for any parameters $a,b,c$, and d.

The following summation formula

$$\begin{array}{c}\hfill \sum _{j=0}^k{\displaystyle \frac{(1-a{q}^{2j})}{(1-a)}}{\displaystyle \frac{{(a,b,c,\frac{a}{bc};q)}_j}{{(q,\frac{aq}{b},\frac{aq}{c},bcq;q)}_j}}{q}^j={\displaystyle \frac{{(aq,bq,cq,\frac{aq}{bc};q)}_k}{{(\frac{aq}{b},\frac{aq}{c},bcq,q;q)}_k}},\end{array}$$

(4)

for q-series with three parameters $a,b,c$ can be obtained by different q-analogue of transformation formulas in the theory of q-series and has been extended by many authors to a more and more general summation formula. See [4,5], for instance. In this paper, we obtain an equivalent identity of (4) which can be proved simply via the telescoping method.

Theorem 2.

Notations are noted as above. We have

$$\begin{array}{cc}\hfill & \sum _{j=1}^k{\displaystyle \frac{{(a,b,c,abc;q)}_j}{{(abq,bcq,caq,q;q)}_j}}{\displaystyle \frac{(abc{q}^{2j}-1)q^j}{1-abc}}=1-{\displaystyle \frac{{(aq,bq,cq,abcq;q)}_k}{{(abq,bcq,caq,q;q)}_k}},\hfill \end{array}$$

(5)

for any parameters $a,b$ and c.

Specifically, our results (i) provide concise closed-form expressions and streamlined proofs for otherwise intricate q-series, (ii) reveal deeper structural links among partition theory, generating functions, and hypergeometric series, and (iii) serve as building blocks for constructing more complex q-analogue in number theory and mathematical physics. In the following section, we will present proofs of Theorems 1 and 2. In Section 3, we will derive double-series representations of several well-known zeta-value constants, thus demonstrating the practical applications of our results.

2. Proofs

Recall that ${\Delta }_j{(a,b;q)}_j={(a,b;q)}_j-{(a,b;q)}_{j+1}$. For any parameters $a,b,c$ and d, the telescoping sum

$$\sum _{j=1}^k{\Delta }_j{\displaystyle \frac{{(a,b;q)}_j}{{(c,d;q)}_j}},$$

is simply equal to

$${\displaystyle \frac{(1-a)(1-b)}{(1-c)(1-d)}}-{\displaystyle \frac{{(a,b;q)}_{k+1}}{{(c,d;q)}_{k+1}}}.$$

Proof of Theorem 1.

We rewrite ${\Delta }_j{\displaystyle \frac{{(a,b;q)}_j}{{(c,d;q)}_j}}={\displaystyle \frac{{(a,b;q)}_j}{{(c,d;q)}_j}}-{\displaystyle \frac{{(a,b;q)}_{j+1}}{{(c,d;q)}_{j+1}}}$ as

$${\displaystyle \frac{{(a,b;q)}_j}{{(c,d;q)}_{j+1}}}\left[(a+b-c-d)q^j+(cd-ab)q^{2j}\right].$$

Therefore, we have

$$\sum _{j=1}^k{\displaystyle \frac{{(a,b;q)}_j}{{(c,d;q)}_{j+1}}}\left[(a+b-c-d)q^j+(cd-ab)q^{2j}\right]={\displaystyle \frac{(1-a)(1-b)}{(1-c)(1-d)}}-{\displaystyle \frac{{(a,b;q)}_{k+1}}{{(c,d;q)}_{k+1}}},$$

or (by dividing ${\displaystyle \frac{(1-a)(1-b)}{(1-c)(1-d)}}$ on both sides)

$$\sum _{j=1}^k{\displaystyle \frac{{(aq,bq;q)}_j}{{(cq,dq;q)}_j}}{\displaystyle \frac{(a+b-c-d)q^j+(cd-ab)q^{2j}}{(1-a{q}^j)(1-b{q}^j)}}=1-{\displaystyle \frac{{(aq,bq;q)}_k}{{(cq,dq;q)}_k}}.$$

Let $f_q\left(i\right)$ be a q-analogue function of i, and we multiply the above identity of both sides by $f_q\left(i\right)$ and then sum over i form 1 to m to obtain the desired identity. □

A special case of Theorem 1 with $b=d=0$ leads to

Corollary 1.

Notations are noted as above. For any parameters $a,c$ and any positive integers $m,n,k$, we have

$$\sum _{i=1}^m\sum _{j=1}^k{\displaystyle \frac{{(aq;q)}_j}{{(cq;q)}_j}}{\displaystyle \frac{(a-c)q^j}{1-a{q}^j}}{f}_q\left(i\right)=\sum _{i=1}^m{f}_q\left(i\right)\left(1-{\displaystyle \frac{{(aq;q)}_k}{{(cq;q)}_k}}\right).$$

(6)

Recall that (the inner summation on the left-hand side of Equation (6))

$$\sum _{j=1}^k{\displaystyle \frac{(a-b)q^j{(aq;q)}_j}{(1-a{q}^j){(bq;q)}_j}}=1-{\displaystyle \frac{{(aq;q)}_k}{{(bq;q)}_k}}.$$

(7)

Let t be a positive integer. Just substituting q with $q^t$, a with $q^{1-t}$ and b with $q^{tni-t+1}$ into Equation (7), we have

$$\sum _{j=1}^k{\displaystyle \frac{q^{tj-t+1}(1-q^{tni}){(q;q^t)}_j}{(1-q^{tj-t+1}){(q^{tni+1};q^t)}_j}}=1-{\displaystyle \frac{{(q;q^t)}_k}{{(q^{tni+1};q^t)}_k}},$$

or

$$\sum _{j=1}^k{\displaystyle \frac{{\left[tni\right]}_q{q}^{tj-t+1}{(q;q^t)}_j}{{[tj-t+1]}_q{(q^{tni+1};q^t)}_j}}=1-{\displaystyle \frac{{(q;q^t)}_k}{{(q^{tni+1};q^t)}_k}}.$$

Now, notice that

$${\displaystyle \frac{{\left[\begin{array}{c}ni+j\\ j\end{array}\right]}_{q^t}}{{\left[\begin{array}{c}tni+tj\\ tj\end{array}\right]}_q}}={\displaystyle \frac{{(q^{tni+t};q^t)}_j{(q;q)}_{tj}}{{(q^t;q^t)}_j{(q^{tni+1};q)}_{tj}}},$$

and

$${\displaystyle \frac{{(q^t;q^t)}_j{(q;q^t)}_j}{{(q;q)}_{tj}}}=\prod _{{\displaystyle \genfrac{}{}{0pt}{}{\ell =1}{\ell \not\equiv 0,1\left(\mathrm{mod}t\right)}}}^{tj}{(1-q^{\ell })}^{-1}.$$

So, we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{{(q;q^t)}_j}{{(q^{tni+1};q^t)}_j}}& ={\displaystyle \frac{{(q^{tni+t};q^t)}_j{(q;q)}_{tj}}{{(q^t;q^t)}_j{(q^{tni+1};q)}_{tj}}}·{\displaystyle \frac{{(q^t;q^t)}_j{(q;q^t)}_j}{{(q;q)}_{tj}}}·{\displaystyle \frac{{(q^{tni+1};q)}_{tj}}{{(q^{tni+1};q^t)}_j{(q^{tni+t};q^t)}_j}}\hfill \\ \hfill & ={\displaystyle \frac{{\left[\begin{array}{c}ni+j\\ j\end{array}\right]}_{q^t}}{{\left[\begin{array}{c}tni+tj\\ tj\end{array}\right]}_q}}\prod _{{\displaystyle \genfrac{}{}{0pt}{}{\ell =1}{\ell \not\equiv 0,1\left(\mathrm{mod}t\right)}}}^{tj}{\displaystyle \frac{1-q^{tni+\ell }}{1-q^{\ell }}},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & \sum _{j=1}^k{\displaystyle \frac{{\left[tni\right]}_q{q}^{tj-t+1}{(q;q^t)}_j}{{[tj-t+1]}_q{(q^{tni+1};q^t)}_j}}\hfill \\ \hfill & =\sum _{j=1}^k{\displaystyle \frac{{\left[\begin{array}{c}ni+j\\ j\end{array}\right]}_{q^t}}{{\left[\begin{array}{c}tni+tj\\ tj\end{array}\right]}_q}}\prod _{{\displaystyle \genfrac{}{}{0pt}{}{\ell =1}{\ell \not\equiv 0,1\left(\mathrm{mod}t\right)}}}^{tj}\left({\displaystyle \frac{1-q^{tni+\ell }}{1-q^{\ell }}}\right)·{\displaystyle \frac{{\left[tni\right]}_q{q}^{tj-t+1}}{{[tj-t+1]}_q}}\hfill \\ \hfill & =1-{\displaystyle \frac{{(q;q^t)}_k}{{(q^{tni+1};q^t)}_k}}.\hfill \end{array}$$

According to Corollary 1, we summarize the above result as a corollary.

Corollary 2.

Notations are noted as before. For any positive integers $m,n,k$ and t, we have

$$\begin{array}{cc}\hfill & \sum _{i=1}^m\sum _{j=1}^k{\displaystyle \frac{{\left[\begin{array}{c}ni+j\\ j\end{array}\right]}_{q^t}}{{\left[\begin{array}{c}tni+tj\\ tj\end{array}\right]}_q}}\prod _{{\displaystyle \genfrac{}{}{0pt}{}{\ell =1}{\ell \not\equiv 0,1\left(\mathrm{mod}t\right)}}}^{tj}\left({\displaystyle \frac{{[tni+\ell ]}_q}{{[\ell ]}_q}}\right)·{\displaystyle \frac{{\left[tni\right]}_q{q}^{tj-t+1}{f}_q\left(i\right)}{{[tj-t+1]}_q}}\hfill \\ \hfill & =\sum _{i=1}^m{f}_q\left(i\right)\left(1-{\displaystyle \frac{{(q;q^t)}_k}{{(q^{tni+1};q^t)}_k}}\right).\hfill \end{array}$$

(8)

Equation (8) can be viewed as a more general formula which appeared in Chen’s paper [3] (Theorem 1). For $t=1,2$, we recover Equation (1) and Equation (2) respectively. For $t=3$, into the above Equation (8), we have

$$\begin{array}{cc}\hfill & \sum _{i=1}^m\sum _{j=1}^k{\displaystyle \frac{{\left[\begin{array}{c}ni+j\\ j\end{array}\right]}_{q^3}}{{\left[\begin{array}{c}3ni+3j\\ 3j\end{array}\right]}_q}}·{\displaystyle \frac{{[3ni+2]}_q{[2ni+5]}_q\cdots {[3ni+3j-1]}_q}{{\left[2\right]}_q{\left[5\right]}_q\cdots {[3j-1]}_q}}·{\displaystyle \frac{{\left[3ni\right]}_q{q}^{3j-2}{f}_q\left(i\right)}{{[3j-2]}_q}}\hfill \\ \hfill & =\sum _{i=1}^m{f}_q\left(i\right)\left(1-{\displaystyle \frac{{(q;q^3)}_k}{{(q^{3ni+1};q^3)}_k}}\right).\hfill \end{array}$$

Assume that $f\left(i\right)={lim}_{q\to 1^{-}}{f}_q\left(i\right)$ exist and the series ${\sum }_{i=1}^{\infty }f\left(i\right)$ converge. Putting $n=1$ and letting $q\to 1^{-}$, and then letting $k\to \infty$ and $m\to \infty$, we have

$$\begin{array}{c}\hfill \sum _{i,j=1}^{\infty }{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{i+j}{j}\right)}{\left(\genfrac{}{}{0pt}{}{3i+3j}{3j}\right)}}·{\displaystyle \frac{(3i+2)(3i+5)\cdots (3i+3j-1)}{2·5\cdots (3j-1)}}·{\displaystyle \frac{3i}{3j-2}}f\left(i\right)=\sum _{i=1}^{\infty }f\left(i\right).\end{array}$$

(9)

Another special case of Theorem 1 with $a=b=d=0$ gives

Corollary 3.

For any parameter c and any positive integers m and k, we have

$$\sum _{i=1}^m\sum _{j=1}^k{\displaystyle \frac{c{q}^j}{{(cq;q)}_j}}{f}_q\left(i\right)=\sum _{i=1}^m{f}_q\left(i\right)\left({\displaystyle \frac{1}{{(cq;q)}_k}}-1\right).$$

We now proceed to present the Proof of Theorem 2 and engage in a detailed discussion of its implications.

Proof of Theorem 2.

We begin our proof with the telescoping sum

$$\sum _{j=1}^k{\Delta }_j{\displaystyle \frac{{(a,b,c,abc;q)}_j}{{(ab,bc,ca;q)}_j{(q;q)}_{j-1}}}.$$

On one hand, this telescoping sum can be easily evaluated as

$${\displaystyle \frac{{(a,b,c,abc;q)}_1}{{(ab,bc,ca;q)}_1}}-{\displaystyle \frac{{(a,b,c,abc;q)}_{k+1}}{{(ab,bc,ca;q)}_{k+1}{(q;q)}_k}}.$$

Note that

$$\begin{array}{cc}\hfill {\Delta }_j& {\displaystyle \frac{{(a,b,c,abc;q)}_j}{{(ab,bc,ca;q)}_j{(q;q)}_{j-1}}}={\displaystyle \frac{{(a,b,c,abc;q)}_j}{{(ab,bc,ca;q)}_{j+1}{(q;q)}_j}}·\hfill \\ \hfill & \left[(1-ab{q}^j)(1-bc{q}^j)(1-ca{q}^j)(1-q^j)-(1-a{q}^j)(1-b{q}^j)(1-c{q}^j)(1-abc{q}^j)\right].\hfill \end{array}$$

By expanding the expression within the bracket in the above formula, it is easy to see that the coefficients of $q^{2j}$ and $q^{4j}$ are both zero.

In light of the identity

$$(1-\alpha )(1-\beta )(1-\gamma )=1-\alpha -\beta -\gamma +\alpha \beta +\beta \gamma +\gamma \alpha -\alpha \beta \gamma ,$$

the inner of the summation can be rewritten as

$${\Delta }_j{\displaystyle \frac{{(a,b,c,abc;q)}_j}{{(ab,bc,ca;q)}_j{(q;q)}_{j-1}}}={\displaystyle \frac{{(a,b,c,abc;q)}_j}{{(ab,bc,ca;q)}_{j+1}{(q;q)}_j}}·\left[q^j(1-a)(1-b)(1-c)(abc{q}^{2j}-1)\right].$$

Dividing both sides by ${\displaystyle \frac{(1-a)(1-b)(1-c)(1-abc)}{(1-ab)(1-bc)(1-ca)}}$ and taking summation on j, we obtain the desired summation formula. □

We remark here that the summation Formula (4) can be derived easily by substituting a with ${\displaystyle \frac{a}{bc}}$ in the summation Formula (5). In fact, these two Formulas (4) and (5) are equivalent, because we replace a with $abc$ in (4), then we get (5). However, our proof of (5) is much simpler and more elementary.

Likewise, we turn our attention to the telescoping sum

$$\sum _{j=1}^k{\Delta }_j{\displaystyle \frac{{(ab,bc,ca;q)}_j{(q;q)}_{j-1}}{{(a,b,c,abc;q)}_j}},$$

which is simply equal to

$${\displaystyle \frac{{(ab,bc,ca;q)}_1}{{(a,b,c,abc;q)}_1}}-{\displaystyle \frac{{(ab,bc,ca;q)}_{k+1}{(q;q)}_k}{{(a,b,c,abc;q)}_{k+1}}}.$$

Then, we obtain the summation formula

$$\begin{array}{c}\hfill \sum _{j=1}^k{\displaystyle \frac{{(abq,bcq,caq,q;q)}_j}{{(aq,bq,cq,abcq;q)}_j}}{\displaystyle \frac{(1-a)(1-b)(1-c)q^j(1-abc{q}^{2j})}{(1-ab{q}^j)(1-bc{q}^j)(1-ca{q}^j)(1-q^j)}}=1-{\displaystyle \frac{{(abq,bcq,caq,q;q)}_k}{{(aq,bq,cq,abcq;q)}_k}}.\end{array}$$

If we let $a=b=c=q^i(i\ge 1)$ in the above formula, we get

$$\begin{array}{c}\hfill \sum _{j=1}^k{\displaystyle \frac{{(q^{2i+1};q)}_j^3{(q;q)}_j}{{(q^{i+1};q)}_j^3{(q^{3i+1};q)}_j}}{\displaystyle \frac{{(1-q^i)}^3{q}^j(1-q^{3i+2j})}{{(1-q^{2i+j})}^3(1-q^j)}}=1-{\displaystyle \frac{{(q^{2i+1};q)}_k^3{(q;q)}_k}{{(q^{i+1};q)}_k^3{(q^{3i+1};q)}_k}},\end{array}$$

or

$$\begin{array}{c}\hfill \sum _{j=1}^k{\displaystyle \frac{{\left[\begin{array}{c}2i+j\\ j\end{array}\right]}_q^3}{{\left[\begin{array}{c}i+j\\ j\end{array}\right]}_q^3{\left[\begin{array}{c}3i+j\\ j\end{array}\right]}_q}}·{\displaystyle \frac{{\left[i\right]}_q^3{[3i+2j]}_q}{{[2i+j]}_q^3{\left[j\right]}_q}}{q}^j=1-{\displaystyle \frac{{\left[\begin{array}{c}2i+k\\ k\end{array}\right]}_q^3}{{\left[\begin{array}{c}i+k\\ k\end{array}\right]}_q^3{\left[\begin{array}{c}3i+k\\ k\end{array}\right]}_q}}.\end{array}$$

By letting $q\to 1^{-}$, we have

$$\begin{array}{c}\hfill \sum _{j=1}^k{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2i+j}{j}\right)}^3}{{\left(\genfrac{}{}{0pt}{}{i+j}{j}\right)}^3\left(\genfrac{}{}{0pt}{}{3i+j}{j}\right)}}·{\displaystyle \frac{i^3(3i+2j)}{{(2i+j)}^{3}j}}=1-{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2i+k}{k}\right)}^3}{{\left(\genfrac{}{}{0pt}{}{i+k}{k}\right)}^3\left(\genfrac{}{}{0pt}{}{3i+k}{k}\right)}}.\end{array}$$

This also implies that

$$\begin{array}{c}\hfill \sum _{i=1}^m\sum _{j=1}^k{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2i+j}{j}\right)}^3}{{\left(\genfrac{}{}{0pt}{}{i+j}{j}\right)}^3\left(\genfrac{}{}{0pt}{}{3i+j}{j}\right)}}·{\displaystyle \frac{i^3(3i+2j)}{{(2i+j)}^{3}j}}f\left(i\right)=\sum _{i=1}^{m}f\left(i\right)\left(1-{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2i+k}{k}\right)}^3}{{\left(\genfrac{}{}{0pt}{}{i+k}{k}\right)}^3\left(\genfrac{}{}{0pt}{}{3i+k}{k}\right)}}\right).\end{array}$$

Here, $f\left(i\right)$ is a function of i, and we assume that the series ${\sum }_{i=1}^{\infty }f\left(i\right)$ converges. Now, fix k and note that

$$\underset{i\to \infty }{lim}{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2i+k}{k}\right)}^3}{{\left(\genfrac{}{}{0pt}{}{i+k}{k}\right)}^3\left(\genfrac{}{}{0pt}{}{3i+k}{k}\right)}}=0.$$

Hence, we let $m\to \infty$ and then let $k\to \infty$ to have

$$\begin{array}{c}\hfill \sum _{i,j=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2i+j-1}{j-1}\right)}^3}{{\left(\genfrac{}{}{0pt}{}{i+j}{j}\right)}^3\left(\genfrac{}{}{0pt}{}{3i+j}{j}\right)}}·{\displaystyle \frac{i^3(3i+2j)}{j^4}}f\left(i\right)=\sum _{i=1}^{\infty }f\left(i\right).\end{array}$$

(10)

3. Conclusions

In this short article, we derive some q-series identities via the telescoping method. In particular, we provide a simple way to express a double sum as a single sum and vice versa. Take, for example, our Corollary 2 or identity (8), which is the general consequence of (1) and (2). In addition, in light of (9) and (10), by setting $f\left(i\right)=i^{-s}$, we see that

$$\begin{array}{c}\hfill \zeta \left(s\right)=\sum _{i,j=1}^{\infty }{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{i+j}{j}\right)}{\left(\genfrac{}{}{0pt}{}{3i+3j}{3j}\right)}}·{\displaystyle \frac{(3i+2)(3i+5)\cdots (3i+3j-1)}{2·5\cdots (3j-1)}}·{\displaystyle \frac{3}{i^{s-1}(3j-2)}}\end{array}$$

and

$$\begin{array}{c}\hfill \zeta \left(s\right)=\sum _{i,j=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2i+j-1}{j-1}\right)}^3}{{\left(\genfrac{}{}{0pt}{}{i+j}{j}\right)}^3\left(\genfrac{}{}{0pt}{}{3i+j}{j}\right)}}·{\displaystyle \frac{3i+2j}{i^{s-3}{j}^4}},\end{array}$$

for real $s>1$, respectively. In particular, we have

$$\zeta \left(3\right)=\sum _{i,j=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2i+j-1}{j-1}\right)}^3}{{\left(\genfrac{}{}{0pt}{}{i+j}{j}\right)}^3\left(\genfrac{}{}{0pt}{}{3i+j}{j}\right)}}·{\displaystyle \frac{3i+2j}{j^4}}.$$

It is well-known that ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{i^2}}{\sum }_{j=1}^i{\displaystyle \frac{1}{j}}=2\zeta \left(3\right)$ and ${\sum }_{i=1}^{\infty }{\displaystyle \frac{1}{i^2}}{\sum }_{k=1}^i{\displaystyle \frac{1}{2k-1}}={\displaystyle \frac{7}{4}}\zeta \left(3\right)$. Moreover, by setting $f\left(i\right)={\displaystyle \frac{H_i}{i^2}}$ in (9), we obtain

$$2\zeta \left(3\right)=\sum _{i=1}^{\infty }{\displaystyle \frac{H_i}{i^2}}=\sum _{i,j=1}^{\infty }{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{i+j}{j}\right)}{\left(\genfrac{}{}{0pt}{}{3i+3j}{3j}\right)}}·{\displaystyle \frac{(3i+2)(3i+5)\cdots (3i+3j-1)}{2·5\cdots (3j-1)}}·{\displaystyle \frac{3H_i}{i(3j-2)}},$$

and, in particular, by setting $f\left(i\right)={\displaystyle \frac{O_i}{i^2}}$ in (10), we obtain

$${\displaystyle \frac{7}{4}}\zeta \left(3\right)=\sum _{i=1}^{\infty }{\displaystyle \frac{O_i}{i^2}}=\sum _{i,j=1}^{\infty }{\displaystyle \frac{{\left(\genfrac{}{}{0pt}{}{2i+j-1}{j-1}\right)}^3}{{\left(\genfrac{}{}{0pt}{}{i+j}{j}\right)}^3\left(\genfrac{}{}{0pt}{}{3i+j}{j}\right)}}·{\displaystyle \frac{i(3i+2j)O_i}{j^4}},$$

where $O_i={\sum }_{k=1}^i{\displaystyle \frac{1}{2k-1}}$.

These identities can be derived in many different ways, and it would be interesting to obtain similar identities using the telescoping method. We invite interested readers to use Theorems 1 and 2 to create new or interesting identities and explore more of these results.