June 01, 2026
Maesaka, Seki and Watanabe proved a formula for multiple harmonic sums. Yamamoto generalized it to Schur-type multiple harmonic sums, and the second author proved a \(q\)-analogue of this generalization. In this paper, we give two applications of the \(q\)-analogue formula. The first is an alternative proof of the duality of a \(q\)-analogue of multiple zeta values. The second is a proof of an identity for a \(q\)-analogue of the Kawashima function.
We call a tuple of positive integers an index. An index \((k_{1}, \ldots , k_{r})\) is said to be admissible if \(k_{r}\ge 2\). For an admissible index \(\boldsymbol{k}=(k_{1}, \ldots , k_{r})\), multiple zeta value (MZV for short) \(\zeta(\boldsymbol{k})\) is defined by \[\begin{align} \zeta(\boldsymbol{k})= \sum_{0<m_{1}<\cdots<m_{r}}\frac{1}{m_{1}^{k_{1}}\cdots m_{r}^{k_{r}}}. \end{align}\] In [1], Maesaka, Seki and Watanabe obtained an interesting formula, which we call the MSW formula in this paper, for the truncated sum \[\begin{align} \zeta_{<N}(\boldsymbol{k})= \sum_{0<m_{1}<\cdots<m_{r}<N} \frac{1}{m_{1}^{k_{1}}\cdots m_{r}^{k_{r}}}. \end{align}\]
Theorem 1. [1] For an index \(\boldsymbol{k}=(k_{1}, \ldots , k_{r})\) and a positive integer \(N\), we set \[\begin{align} \zeta_{<N}^{\flat}(\boldsymbol{k})= \sum \prod_{j=1}^{r} \left( \frac{1}{N-n_{j,1}} \prod_{l=2}^{k_{j}}\frac{1}{n_{j,l}} \right), \end{align}\] where the sum is taken over all integers \(n_{j, l} \, (1\le j \le r, \, 1\le l \le k_{j})\) satisfying \[\begin{align} 0<n_{j,1}\leq\cdots\leq n_{j,k_{j}}<N \quad (1\le j \le r), \qquad n_{j,k_{j}}<n_{(j+1),1}\quad (1\leq j<r). \label{eq:sum-condition-MSW} \end{align}\qquad{(1)}\] Then, for any index \(\boldsymbol{k}\) and any positive integer \(N\), it holds that \[\begin{align} \zeta_{<N}(\boldsymbol{k})=\zeta_{<N}^{\flat}(\boldsymbol{k}). \label{eq:MSW} \end{align}\qquad{(2)}\]
For example, \(\zeta^{\flat}_{<N}(1,2,3)\) is written as \[\begin{align} \zeta^{\flat}_{<N}(1,2,3)=\sum_{0<n_{1}<n_{2}\leq n_{3}<n_{4}\leq n_{5}\leq n_{6}<N} \frac{1}{(N-n_{1})(N-n_{2})n_{3}(N-n_{4})n_{5}n_{6}}. \end{align}\] In the limit as \(N\to \infty\), the sum converges to the multiple integral \[\begin{align} \idotsint\limits_{0<t_{1}<t_{2}<t_{3}<t_{4}<t_{5}<t_{6}<1} \frac{\mathrm{d}t_{1}}{1-t_{1}}\frac{\mathrm{d}t_{2}\mathrm{d}t_{3}}{(1-t_{2})t_{3}}\frac{\mathrm{d}t_{4}\mathrm{d}t_{5}\mathrm{d}t_{6}}{(1-t_{4})t_{5}t_{6}}, \end{align}\] which is the iterated integral representation of the MZV \(\zeta(1, 2, 3)\). In this sense, the MSW formula ?? gives a finite discretization of the iterated integral representation of MZVs.
In [2], Yamamoto generalized the MSW formula ?? to Schur-type multiple harmonic sums, which may be viewed as a finite discretization of the integral formula for Schur multiple zeta values established by Hirose, Murahara and Onozuka [3]. It contains the star-version of the MSW formula:
Theorem 2. [2] For an index \(\boldsymbol{k}=(k_{1}, \ldots , k_{r})\) and a positive integer \(N\), we set \[\begin{align} \zeta_{<N}^{\star}(\boldsymbol{k})= \sum_{0<m_{1}\le \cdots \le m_{r}<N} \frac{1}{m_{1}^{k_{1}} \cdots m_{r}^{k_{r}}} \end{align}\] and \[\begin{align} \zeta_{<N}^{\star \flat}(\boldsymbol{k})= \sum \prod_{j=1}^{r} \left( \frac{1}{N-n_{j,1}} \prod_{l=2}^{k_{j}}\frac{1}{n_{j,l}} \right), \end{align}\] where the sum is taken over all integers \(n_{j, l} \, (1\le j \le r, \, 1\le l \le k_{j})\) satisfying \[\begin{align} 0<n_{j,1}\leq\cdots\leq n_{j,k_{j}}<N \quad (1\le j \le r), \qquad n_{j,k_{j}} \ge n_{j-1,1} \quad (1<j\le r). \label{eq:sum-condition-star-MSW} \end{align}\qquad{(3)}\] Then, for any index \(\boldsymbol{k}\) and any positive integer \(N\), it holds that \[\begin{align} \zeta_{<N}^{\star}(\boldsymbol{k})= \zeta_{<N}^{\star \flat}(\boldsymbol{k}). \label{eq:MSW-star} \end{align}\qquad{(4)}\]
In [4], the second author obtained a \(q\)-analogue of Yamamoto’s generalization. It contains a \(q\)-analogue of ?? and ?? given as follows.
We assume throughout this paper that \(q\) is a constant satisfying \[\begin{align} 0<q<1. \end{align}\] For an integer \(m\), the \(q\)-integer \([m]\) is defined by \([m]=(1-q^{m})/(1-q)\).
For an index \(\boldsymbol{k}=(k_{1}, \ldots , k_{r})\), we set \[\begin{align} \zeta_{<N}^{q}(\boldsymbol{k})= \sum_{0<m_{1}<\cdots<m_{r}<N} \frac{q^{(k_{1}-1)m_{1}+\cdots+(k_{r}-1)m_{r}}}{[m_{1}]^{k_{1}}\cdots [m_{r}]^{k_{r}}} \label{eq:q-flat-sum} \end{align}\tag{1}\] and define its star-version \(\zeta_{<N}^{\star, q}(\boldsymbol{k})\) as 1 with the summation range replaced by \(0<m_{1}\le \cdots\le m_{r}<N\). Note that \(\zeta_{<N}^{q}(\boldsymbol{k})\) is a truncated sum of the Bradley-Zhao model of a \(q\)-analogue of MZV (\(q\)MZV for short) defined by \[\begin{align} \zeta^{q}(\boldsymbol{k})= \sum_{0<m_{1}<\cdots<m_{r}}\frac{q^{(k_{1}-1)m_{1}+\cdots+(k_{r}-1)m_{r}}}{[m_{1}]^{k_{1}}\cdots [m_{r}]^{k_{r}}} \label{eq:qMZV} \end{align}\tag{2}\] for an admissible index \(\boldsymbol{k}=(k_{1}, \ldots, k_{r})\).
Theorem 3. [4] For an index \(\boldsymbol{k}=(k_{1}, \ldots , k_{r})\) and a positive integer \(N\), we set \[\begin{align} \zeta_{<N}^{q\flat}(\boldsymbol{k})= \sum\prod_{j=1}^{r} \left( \frac{1}{[N-n_{j,1}]} \prod_{l=2}^{k_{j}} \frac{q^{n_{j,l}}}{[n_{j,l}]} \right), \label{eq:q-flat} \end{align}\qquad{(5)}\] where the sum is taken over all integers \(n_{j, l} \, (1\le j \le r, \, 1\le l \le k_{j})\) satisfying ?? . We also define its star-version \(\zeta_{<N}^{\star, q\flat}(\boldsymbol{k})\) as ?? with the summation range replaced by ?? . Then, for any index \(\boldsymbol{k}\) and any positive integer \(N\), we have \[\begin{align} \zeta^{q}_{<N}(\boldsymbol{k}) & =\zeta^{q\flat}_{<N}(\boldsymbol{k}), \label{eq:qMSW} \\ \zeta^{\star, q}_{<N}(\boldsymbol{k}) & =\zeta^{\star, q\flat}_{<N}(\boldsymbol{k}). \label{eq:star-qMSW} \end{align}\] {#eq: sublabel=eq:eq:qMSW,eq:eq:star-qMSW}
The purpose of this paper is to discuss two applications of the formulas ?? and ?? . The first is an alternative proof for the duality of \(q\)MZVs 2 using the formula ?? . The second is a proof of an identity of a \(q\)-analogue of the Kawashima function using the formula ?? .
In Section 2, we present a new proof of the duality of \(q\)MZVs.
Let \(\boldsymbol{k}\) be an admissible index. There uniquely exist positive integers \(s\) and \(a_{1}, \ldots , a_{s}, b_{1}, \ldots , b_{s}\) such that \[\begin{align} \boldsymbol{k}=(\underbrace{1,\ldots,1}_{a_{1}-1},b_{1}+1,\ldots,\underbrace{1,\ldots,1}_{a_{s}-1},b_{s}\color{black}+1). \label{eq:def-ab} \end{align}\tag{3}\] Then the dual index \(\boldsymbol{k}^{\dagger}\) of \(\boldsymbol{k}\) is defined by \[\begin{align} \boldsymbol{k}^{\dagger}=(\underbrace{1,\ldots,1}_{b_{s}-1},a_{s}+1,\ldots,\underbrace{1,\ldots,1}_{b_{1}-1},a_{1}+1). \end{align}\]
Theorem 4 (Duality of \(q\)MZVs, [5]). For any admissible index \(\boldsymbol{k}\), it holds that \(\zeta^{q}(\boldsymbol{k})=\zeta^{q}(\boldsymbol{k}^{\dagger})\).
By taking the limit as \(q\to 1\), we recover the duality \(\zeta(\boldsymbol{k})=\zeta(\boldsymbol{k}^{\dagger})\) of MZVs.
The paper [1] provides a proof of the duality of MZVs by means of the MSW formula. It is proved that, for any admissible index \(\boldsymbol{k}\), there exists \(J>0\) such that \[\begin{align} \zeta_{<N}(\boldsymbol{k})-\zeta_{<N}(\boldsymbol{k}^{\dagger})= N^{-1}O((\log{N})^{J}) \qquad (N \to \infty). \end{align}\] By taking the limit as \(N\to \infty\), we obtain the duality \(\zeta(\boldsymbol{k})=\zeta(\boldsymbol{k}^{\dagger})\).
In contrast to the MZV case, our proof of the duality of \(q\)MZVs proceeds as follows. Using the formula ?? , one can obtain, for any admissible index \(\boldsymbol{k}\), a finite sum \(\phi_{<N}^{q}(\boldsymbol{k})\) such that \[\begin{align} \zeta_{<N}^{q}(\boldsymbol{k})=\zeta_{<N}^{q\flat}(\boldsymbol{k})=\phi_{<N}^{q}(\boldsymbol{k})+q^{N}O(N^{J}) \end{align}\] with some \(J>0\) and \[\begin{align} \phi_{<N}^{q}(\boldsymbol{k}) \to \zeta^{q}(\boldsymbol{k}^{\dagger}) \label{eq:phi-zeta} \end{align}\tag{4}\] as \(N\to \infty\). The duality is then derived from this. To show 4 , we employ the same technique as was used to prove the resummation identity given in [6].
In Section 3, we prove an identity of a \(q\)-analogue of the Kawashima function.
In [7], Kawashima introduced a function \(F_{\boldsymbol{k}}(z)\) for an index \(\boldsymbol{k}\), which we call the Kawashima function. The function \(F_{\boldsymbol{k}}(z)\) interpolates the truncated sum \(\zeta_{<N}(\boldsymbol{k})\) as \(F_{\boldsymbol{k}}(N-1)=\zeta_{<N}(\boldsymbol{k})\) for any positive integer \(N\). In [8], Kawashima obtained an alternative expression \(G_{\boldsymbol{k}}(z)\) of the Kawashima function such that \[\begin{align} F_{\boldsymbol{k}}(z)=G_{\overleftarrow{\boldsymbol{k}}}(z) \label{eq:Kawashima-identity} \end{align}\tag{5}\] in the region \(\mathop{\mathrm{Re}}{z}>-1\), where \(\overleftarrow{\boldsymbol{k}}:=(k_{r}, k_{r-1}, \ldots , k_{1})\) for \(\boldsymbol{k}=(k_{1}, \ldots , k_{r})\). The key to the proof of 5 is the identity \[\begin{align} F_{\boldsymbol{k}}(N-1)=G_{\overleftarrow{\boldsymbol{k}}}(N-1) \label{eq:key-identity} \end{align}\tag{6}\] for any positive integer \(N\). In [2], Yamamoto pointed out that the identity 6 is nothing but the star-version ?? of the MSW formula.
In [9], the first author defined a \(q\)-analogue \(F_{\boldsymbol{k}}^{q}(z)\) of the Kawashima function. In Section 3, we introduce a \(q\)-analogue \(G_{\boldsymbol{k}}^{q}(z)\) of the function \(G_{\boldsymbol{k}}(z)\) and show the identity \(F_{\boldsymbol{k}}^{q}(z)=G_{\overleftarrow{\boldsymbol{k}}}^{q}(z)\) on the disk \(|z|<q^{-1}\), which follows from the formula ?? and the identity theorem for analytic functions.
Throughout this paper we set \(|\boldsymbol{k}|=\sum_{a=1}^{r}k_{a}\) for an index \(\boldsymbol{k}=(k_{1}, \ldots , k_{r})\).
This work was supported by JSPS KAKENHI Grant Number JP22K03243.
In this section, we give an alternative proof of Theorem 4.
Proposition 5. For an index \(\boldsymbol{k}=(k_{1}, \ldots , k_{r})\) and a positive integer \(N\), we set \[\begin{align} \phi_{<N}^{q}(\boldsymbol{k})=(1-q)^{r} \sum \prod_{j=1}^{r} \left( \prod_{l=2}^{k_{j}}\frac{q^{n_{j,l}}}{[n_{j,l}]} \right), \end{align}\] where the sum is taken over all integers \(n_{j, l} \, (1\le j \le r, \, 1\le l \le k_{j})\) satisfying ?? . Then, for any admissible index \(\boldsymbol{k}\), it holds that \[\begin{align} \label{eq:32asymptotic1} \zeta_{<N}^{q\flat}(\boldsymbol{k})=\phi_{<N}^{q}(\boldsymbol{k})+q^{N}O(N^{J}) \qquad (N \to \infty) \end{align}\qquad{(6)}\] for some \(J>0\) independent of \(N\).
Proof. Suppose that \(\boldsymbol{k}=(k_{1}, \ldots , k_{r})\) is admissible. Note that \(k_{r}\ge 2\). Since \(0<q<1\), we have \(1/[m]\le 1\) for any positive integer \(m\). Therefore, if \(0<n_{j,1}\le n_{r, k_{r}}<N\), it holds that \[\begin{align} \label{eq:32key32tool} 0<\frac{1}{[N-n_{j,1}]}\frac{q^{n_{r, k_{r}}}}{[n_{r, k_{r}}]}= \left(1-q+\frac{q^{N-n_{j,1}}}{[N-n_{j,1}]}\right) \frac{q^{n_{r, k_{r}}}}{[n_{r, k_{r}}]} \le (1-q)\frac{q^{n_{r, k_{r}}}}{[n_{r, k_{r}}]}+q^{N}. \end{align}\tag{7}\] Applying 7 to \(\zeta^{q\flat}_{<N}(\boldsymbol{k})\) for \(1\le j\le r\), we obtain ?? . ◻
Lemma 6. Suppose that \(a, b\ge 1\) and \(c, d\ge 0\). It holds that \[\begin{align} \label{eq:duality-lem} (1-q)^{a}\sum_{d<n_{1}\le \cdots \le n_{b}} \binom{n_{1}-d}{a} \left( \prod_{j=1}^{b-1}\frac{q^{n_{j}}}{[n_{j}]} \right) \frac{q^{n_{b}}}{[n_{b}]}q^{cn_{b}}= \sum_{c<m_{1}<\cdots <m_{b}} \left( \prod_{j=1}^{b-1}\frac{1}{[m_{j}]} \right) \frac{q^{am_{b}}}{[m_{b}]^{a+1}} q^{dm_{b}}. \end{align}\qquad{(7)}\]
Proof. We denote the left-hand side by \(I\). We have \[\begin{align} I=(1-q)^{a+b}\sum_{d\le n_{1}\le \cdots \le n_{b}} \sum_{l_{1}, \ldots, l_{b}\ge 1} \binom{n_{1}-d}{a} q^{\sum_{j=1}^{b}n_{j}l_{j}+cn_{b}} \end{align}\] because \(\binom{n_{1}-d}{a}=0\) if \(n_{1}=d\). We perform the change of summation variables \((n_{1}, \ldots , n_{b}, l_{1}, \ldots , l_{b}) \mapsto (s_{1}, \ldots , s_{b}, m_{1}, \ldots , m_{b})\) defined by \[\begin{align} s_{1}=n_{1}-d, \qquad s_{j}=n_{j}-n_{j-1} \quad (2\le j \le b) \end{align}\] and \[\begin{align} m_{j}=c+\sum_{i=b-j+1}^{b}l_{i} \quad (1\le j \le b). \end{align}\] The sum is then taken over integers \(s_{j}\) and \(m_{j} \, (1\le j \le b)\) satisfying \(s_{1}, \ldots , s_{b}\ge 0\) and \(c<m_{1}<\cdots <m_{b}\). It holds that \[\begin{align} \sum_{j=1}^{b}n_{j}l_{j}+cn_{b}=\sum_{j=1}^{b}s_{j}m_{b-j+1}+dm_{b}. \end{align}\] We take the sum over \(s_{1}, \ldots , s_{b}\ge 0\). By using \[\begin{align} \sum_{s\ge 0}\binom{s}{a}x^{s}=\frac{x^{a}}{(1-x)^{a+1}} \qquad (|x|<1), \end{align}\] we see that \[\begin{align} I=(1-q)^{a+b}\sum_{c<m_{1}<\cdots <m_{b}} \left(\prod_{j=1}^{b-1}\frac{1}{1-q^{m_{j}}}\right) \frac{q^{am_{b}}}{(1-q^{m_{b}})^{a+1}}\, q^{dm_{b}}, \end{align}\] which is equal to the right-hand side of ?? . ◻
Now, we are ready to prove Theorem 4.
Proof of Theorem 4. Let \(\boldsymbol{k}\) be an admissible index. We define positive integers \(a_{j}, b_{j} \, (1\le j \le s)\) by 3 . From the definition of \(\phi_{<N}^{q}(\boldsymbol{k})\), we see that \[\begin{align} \label{eq:phi-limit} \lim_{N\to \infty}\phi_{<N}^{q}(\boldsymbol{k})=(1-q)^{\sum_{j=1}^{s}a_{j}} \sum \prod_{j=1}^{s}\left( \binom{n_{j, 1}-n_{j-1,b_{j-1}}}{a_{j}} \prod_{l=1}^{b_{j}}\frac{q^{n_{j,l}}}{[n_{j,l}]} \right), \end{align}\tag{8}\] where \(n_{0, b_{0}}=0\) and the sum is taken over all integers \(n_{j, l} \, (1\le j\le s, \, 1\le l\le b_{j})\) satisfying \[\begin{align} 0<n_{1, 1}\le \cdots \le n_{1, b_{1}}< n_{2,1}\le \cdots \le n_{2, b_{2}}<\cdots <n_{s, 1}\le \cdots \le n_{s, b_{s}}. \end{align}\] By using ?? repeatedly, we see that the right-hand side of 8 is equal to \(\zeta^{q}(\boldsymbol{k}^{\dagger})\). ◻
For an index \(\boldsymbol{k}\), we define its Hoffman dual \(\boldsymbol{k}^{\vee}\) as follows. We write \(\boldsymbol{k}\) in the form \(\boldsymbol{k}=(1 \square 1 \square \cdots \square 1)\), where \(\square\) is either \(+\) (plus symbol) or \(,\) (comma). Then \(\boldsymbol{k}^{\vee}\) is the index obtained by changing \(+\) to \(,\) and vice versa. For example, if \(\boldsymbol{k}=(2, 1, 3, 2)=(1+1, 1, 1+1+1, 1+1)\), then \(\boldsymbol{k}^{\vee}=(1, 1+1+1,1,1+1,1)=(1,3,1,2,1)\).
Let \(\boldsymbol{k}\) be an index. We set \(\boldsymbol{k}^{\vee}=(k_{1}', \ldots , k_{s}')\) and define the function \(F_{\boldsymbol{k}}^{q}(z)\), which is a \(q\)-analogue of the Kawashima function, by \[\begin{align} F_{\boldsymbol{k}}^{q}(z)=-\sum_{0<m_{1}\le \cdots \le m_{s}} \left( \prod_{a=1}^{s-1}\frac{q^{(k_{a}'-1)m_{a}}}{[m_{a}]^{k_{a}'}} \right) \frac{q^{k_{s}'m_{s}}}{[m_{s}]^{k_{s}'}} \prod_{j=1}^{m_{s}}\frac{z-q^{j-1}}{1-q^{j}}. \label{eq:q-Kawashima-F} \end{align}\tag{9}\]
Proposition 7. The infinite sum in 9 converges absolutely and defines an analytic function in the region \(|z|<q^{-1}\).
Proof. Fix a constant \(c\) with \(0<c<q^{-1}\). Suppose that \(|z|\le c\). Set \(d=q(1-qc)/(1+q)\). Note that \(0<d<1\). We take a positive integer \(M\) such that \(q^{M}<d\). Then, if \(j>M\), it holds that \[\begin{align} \left|\frac{z-q^{j-1}}{1-q^{j}}\right| & \le \left|\frac{z-q^{j-1}}{1-q^{j}}-z\right|+|z|= \frac{q^{j-1}}{1-q^{j}}|1-zq|+|z| \\ & \le \frac{q^{M}}{1-q^{M}}(1+c)+c\le \frac{c+d}{1-d}. \end{align}\] For \(1\le j\le M\), we have \[\begin{align} \left|\frac{z-q^{j-1}}{1-q^{j}}\right| \le \frac{c+1}{1-q}. \end{align}\] For any positive integer \(m\), we have \(1/[m]\le 1\). Therefore, if \(0<m_{1}\le \cdots \le m_{s}\) and \(m_{s}>M\), it holds that \[\begin{align} \left| \left( \prod_{a=1}^{s-1}\frac{q^{(k_{a}'-1)m_{a}}}{[m_{a}]^{k_{a}'}} \right) \frac{q^{k_{s}'m_{s}}}{[m_{s}]^{k_{s}'}} \prod_{j=1}^{m_{s}}\frac{z-q^{j-1}}{1-q^{j}} \right|\le q^{m_{s}} \left(\frac{c+1}{1-q}\right)^{M} \left(\frac{c+d}{1-d}\right)^{m_{s}-M} \end{align}\] because \(k_{s}'\ge 1\). From the definition of \(d\) and \(0<c<q^{-1}\), we see that \[\begin{align} \frac{c+d}{1-d}=(1+q)\frac{1+c}{1+q^{2}c}-1<q^{-1}. \end{align}\] Hence, we can apply the Weierstrass \(M\)-test to the series in 9 . ◻
The function \(F_{\boldsymbol{k}}^{q}(z)\) has the following property.
Proposition 8. [9]For any index \(\boldsymbol{k}=(k_{1}, \ldots , k_{r})\) and any positive integer \(N\), it holds that \[\begin{align} F_{\boldsymbol{k}}^{q}(q^{N-1})=\sum_{0<m_{1}\le \cdots \le m_{r}<N} \frac{q^{m_{1}+\cdots +m_{r}}}{[m_{1}]^{k_{1}} \cdots [m_{r}]^{k_{r}}}. \end{align}\]
For an integer \(m\), we set \[\begin{align} [m+\log_{q}{z}]=\frac{1-zq^{m}}{1-q}. \end{align}\] Note that, if \(z=q^{N}\) for an integer \(N\), we have \([m+\log_{q}{z}]=[m+N]\).
For an index \(\boldsymbol{k}=(k_{1}, \ldots , k_{r})\), we define the function \(G_{\boldsymbol{k}}^{q}(z)\) by \[\begin{align} \label{eq:q-Kawashima-G} G^{q}_{\boldsymbol{k}}(z)=\sum & \left\{ \prod_{j=1}^{r-1} \left( \prod_{l=1}^{k_{a}-1}\frac{1}{[m_{j, l}+\log_{q}{z}]} \, \right) \frac{1}{[m_{j, k_{j}}]} \right\} \\ \nonumber & \times \left( \prod_{l=1}^{k_{r}-1}\frac{1}{[m_{r, l}+\log_{q}{z}]} \right) \left( \frac{q^{m_{r,k_{r}}}}{[m_{r,k_{r}}]}-\frac{q^{m_{r,k_{r}}}z}{[m_{r,k_{r}}+\log_{q}{z}]} \right), \end{align}\tag{10}\] where the sum is over the region \[\begin{align} 0<m_{1,1}\leq\cdots\leq m_{1,k_{1}}<m_{2,1}\leq\cdots\leq m_{2,k_{2}}< \cdots <m_{r,1}\leq\cdots\leq m_{r,k{r}}. \end{align}\]
Proposition 9. The infinite sum in 10 converges absolutely and defines an analytic function in the region \(|z|<q^{-1}\).
Proof. Let \(c\) be a constant satisfying \(0<c<q^{-1}\). Suppose that \(|z|\le c\). For any positive integer \(m\), it holds that \(1/[m]\le 1\) and \[\begin{align} \left|[m+\log_{q}{z}]\right|\ge \frac{1-cq}{1-q}>0. \end{align}\] Hence, we have \[\begin{align} \left| \frac{q^{m}}{[m]}-\frac{q^{m}z}{[m+\log_{q}{z}]} \right| \le q^{m}\left(1+\frac{c(1-q)}{1-cq}\right). \end{align}\] Therefore, the Weierstrass \(M\)-test applies to the series in 10 . ◻
We prove the following lemma, which will be needed to evaluate \(G_{\boldsymbol{k}}(z)\) at \(z=q^{N-1}\) for a positive integer \(N\).
Lemma 10. Let \(N\) be a positive integer. For any positive integers \(m', n', n''\) satisfying \(n'\le n''\le N-1\), the following identities hold. \[\begin{align} \label{eq:32G-flat40141} \frac{1}{[m^{\prime}+N-1]} & \sum_{m=m^{\prime}}^{\infty} \left( \frac{q^{m+N-1-n^{\prime\prime}}}{[m+N-1-n^{\prime\prime}]}- \frac{q^{m+N-n^{\prime}}}{[m+N-n^{\prime}]} \right) \\ & =\sum_{n=n^{\prime}}^{n^{\prime\prime}} \left( \frac{q^{m^{\prime}+N-1-n}}{[m^{\prime}+N-1-n]}- \frac{q^{m^{\prime}+N-1}}{[m^{\prime}+N-1]} \right) \frac{1}{[n]}, \nonumber \\ \label{eq:32G-flat40241} \frac{1}{[m^{\prime}]} & \sum_{m=m^{\prime}+1}^{\infty} \left( \frac{q^{m+N-1-n^{\prime\prime}}}{[m+N-1-n^{\prime\prime}]}- \frac{q^{m+N-n^{\prime}}}{[m+N-n^{\prime}]} \right) \\ & =\sum_{n=n^{\prime}}^{n^{\prime\prime}} \left( \frac{q^{m^{\prime}}}{[m^{\prime}]}- \frac{q^{m^{\prime}+N-n}}{[m^{\prime}+N-n]} \right) \frac{q^{N-n}}{[N-n]}. \nonumber \end{align}\] {#eq: sublabel=eq:eq:32G-flat40141,eq:eq:32G-flat40241}
Proof. Here we prove ?? . The proof of ?? is similar. Since the sum \(\sum_{m\ge 1}q^{m}/[m]\) converges and \(n'\le n''\), we see that the left-hand side is equal to \[\begin{align} \frac{1}{[m'+N-1]}\sum_{n=n'}^{n''}\frac{q^{m'+N-n-1}}{[m'+N-n-1]}. \end{align}\] Using the partial fraction decomposition \[\begin{align} \frac{q^{m'+N-n-1}}{[m'+N-1][m'+N-n-1]}= \left( \frac{q^{m^{\prime}+N-1-n}}{[m^{\prime}+N-1-n]}- \frac{q^{m^{\prime}+N-1}}{[m^{\prime}+N-1]} \right) \frac{1}{[n]}, \end{align}\] we obtain ?? . ◻
Proposition 11. For any index \(\boldsymbol{k}=(k_{1}, \ldots , k_{r})\) and any positive integer \(N\), we have \[\begin{align} G^{q}_{\overset{\leftarrow}{\boldsymbol{k}}}(q^{N-1})= \sum\prod_{j=1}^{r} \left( \frac{q^{N-n_{j,1}}}{[N-n_{j,1}]} \prod_{l=2}^{k_{j}} \frac{1}{[n_{j,l}]} \right), \label{eq:G-value} \end{align}\qquad{(8)}\] where the sum is taken over all integers \(n_{j, l} \, (1\le j \le r, \, 1\le l \le k_{j})\) satisfying ?? .
Proof. We prove the case \(\boldsymbol{k}=(2,3)\) as an illustration. The general case follows similarly. We have \[\begin{align} G^{q}_{(3,2)}(q^{N-1})=\sum_{\substack{ 0<m_{1}\leq m_{2}\leq m_{3}\\ <m_{4}\leq m_{5}}} & \frac{1}{[m_{1}+N-1][m_{2}+N-1][m_{3}][m_{4}+N-1]} \\ & \times\left( \frac{q^{m_{5}}}{[m_{5}]}-\frac{q^{m_{5}+N-1}}{[m_{5}+N-1]} \right). \end{align}\] Applying ?? to the sum over \(m_{5}\), we obtain
\[\begin{align} \label{eq:32G-flat1} G^{q}_{(3,2)}(q^{N-1})= \sum_{0<n_{5}<N}\frac{1}{[n_{5}]}\sum_{0<m_{1}\leq m_{2}\leq m_{3}<m_{4}} & \frac{1}{[m_{1}+N-1][m_{2}+N-1][m_{3}]} \\ & \times\left( \frac{q^{m_{4}+N-1-n_{5}}}{[m_{4}+N-1-n_{5}]}-\frac{q^{m_{4}+N-1}}{[m_{4}+N-1]} \right). \end{align}\tag{11}\] Using ?? to the sum over \(m_{4}\), we get
\[\begin{align} \label{eq:32G-flat2} G^{q}_{(3,2)}(q^{N-1})= & \sum_{0<n_{4}\leq n_{5}<N} \frac{q^{N-n_{4}}}{[N-n_{4}][n_{5}]} \\ & \qquad {}\times \sum_{0<m_{1}\leq m_{2}\leq m_{3}} \frac{1}{[m_{1}+N-1][m_{2}+N-1]} \left( \frac{q^{m_{3}}}{[m_{3}]}-\frac{q^{m_{3}+N-n_{4}}}{[m_{3}+N-n_{4}]} \right). \end{align}\tag{12}\] Applying ?? twice and then ?? , we see that the right-hand side is equal to \[\begin{align} & \sum_{\substack{0<n_{3}<N, \, 0<n_{4}\le n_{5}<N \\ n_{3}\ge n_{4}}} \frac{1}{[n_{3}]}\frac{q^{N-n_{4}}}{[N-n_{4}][n_{5}]} \sum_{0<m_{1}\leq m_{2}}\frac{1}{[m_{1}+N-1]}\left( \frac{q^{m_{2}+N-1-n_{3}}}{[m_{2}+N-1-n_{3}]}-\frac{q^{m_{2}+N-1}}{[m_{2}+N-1]} \right) \\ & =\sum_{\substack{ 0<n_{2}\leq n_{3}<N, \,0<n_{4}\leq n_{5}<N \\ n_{3}\ge n_{4}}} \frac{1}{[n_{2}][n_{3}]}\frac{q^{N-n_{4}}}{[N-n_{4}][n_{5}]} \sum_{0<m_{1}}\left( \frac{q^{m_{1}+N-1-n_{2}}}{[m_{1}+N-1-n_{2}]}-\frac{q^{m_{1}+N-1}}{[m_{1}+N-1]} \right) \\ & =\sum_{\substack{ 0<n_{1}\leq n_{2}\leq n_{3}, \, 0<n_{4}\leq n_{5}<N \\ n_{3}\ge n_{4}}} \frac{q^{N-n_{1}}}{[N-n_{1}][n_{2}][n_{3}]}\frac{q^{N-n_{4}}}{[N-n_{4}][n_{5}]}. \end{align}\] It is equal to the right-hand side of ?? with \(\boldsymbol{k}=(2,3)\). ◻
We now prove the main theorem of this section.
Theorem 12. For any index \(\boldsymbol{k}\), it holds that \[\begin{align} F_{\boldsymbol{k}}(z)=G_{\overleftarrow{\boldsymbol{k}}}(z) \label{eq:q-F61G} \end{align}\qquad{(9)}\] in the region \(|z|<q^{-1}\).
Proof. Let \(N\) be a positive integer. Note that replacing \(q\) by \(q^{-1}\) sends \([m]\) to \(q^{1-m}[m]\) for any integer \(m\). From Proposition 8, we see that \[\begin{align} q^{-|\boldsymbol{k}|}\, F_{\boldsymbol{k}}^{q}(q^{N-1})= \zeta_{<N}^{\star, q}(\boldsymbol{k})\bigg|_{q\to q^{-1}}. \end{align}\] Similarly, from Proposition 11, we find that \[\begin{align} q^{-|\boldsymbol{k}|}\, G_{\overleftarrow{\boldsymbol{k}}}^{q}(q^{N-1})= \zeta_{<N}^{\star, q\flat}(\boldsymbol{k})\bigg|_{q\to q^{-1}}. \end{align}\] Since the formula ?? is an identity of rational functions in \(q\), it implies that \(F_{\boldsymbol{k}}^{q}(q^{N-1})=G_{\overleftarrow{\boldsymbol{k}}}^{q}(q^{N-1})\) for any positive integer \(N\). Now the desired equality follows from the identity theorem for analytic functions. ◻
Remark 13. In the case \(\boldsymbol{k}=(1)\), the identity ?? can be proved directly by means of the \(q\)-Gauss summation formula, as follows.
We set \((z)_{m}=\prod_{j=0}^{m-1}(1-q^{j}z)\) for \(m \ge 0\) and \((z)_{\infty}=\prod_{j=0}^{\infty}(1-q^{j}z)\). It holds that \[\begin{align} F_{(1)}^{q}(z)=-(1-q)\sum_{m\ge 1} \frac{1}{1-q^{m}}\frac{(z^{-1})_{m}}{(q)_{m}} (qz)^{m}. \end{align}\] Since \[\begin{align} \frac{1}{1-q^{m}}=\lim_{a\to 1}\frac{1}{1-a}\frac{(a)_{m}}{(aq)_{m}} \end{align}\] for \(m\ge 1\), we find that \[\begin{align} F_{(1)}^{q}(z)=(1-q)\lim_{a\to 1}\frac{1}{a-1} \left( {}_{2}\phi_{1}(a, z^{-1}, aq; qz)-1\right), \end{align}\] where \({}_{2}\phi_{1}(a, b, c; z)\) is the \(q\)-hypergeometric series \[\begin{align} {}_{2}\phi_{1}(a, b, c; z)=\sum_{m\ge 0} \frac{(a)_{m}(b)_{m}}{(c)_{m}(q)_{m}}z^{m}. \end{align}\] The \(q\)-Gauss summation formula \[\begin{align} {}_{2}\phi_{1}(a, b, c; c/ab)= \frac{(c/a)_{\infty}(c/b)_{\infty}}{(c)_{\infty}(c/ab)_{\infty}} \qquad (|c/ab|<1) \end{align}\] implies that \[\begin{align} F_{(1)}^{q}(z)=(1-q)\lim_{a\to 1}\frac{1}{a-1} \left( g(a; z)-1\right), \end{align}\] where \[\begin{align} g(a; z)=\frac{(q)_{\infty}(aqz)_{\infty}}{(aq)_{\infty}(qz)_{\infty}}. \end{align}\] Since \(g(1; z)=1\), we have \[\begin{align} F_{(1)}^{q}(z) & =(1-q)\frac{\partial}{\partial a}g(a; z)\bigg|_{a=1} =(1-q)\frac{\partial}{\partial a}\left(\log{g(a; z)}\right)\bigg|_{a=1} \\ & =(1-q)\sum_{m\ge 1}\left(\frac{q^{m}}{1-q^{m}}-\frac{q^{m}z}{1-q^{m}z}\right)= G_{(1)}^{q}(z). \end{align}\]