# Evaluation of q-Pochhammer series

Let $f_q(x)$ be the generating function of the sequence $(q;q)_n$: $$f_q(x):=\sum_{n=0}^{\infty} (q;q)_n x^n,$$ where $(q;q)_n: = (1-q)(1-q^2) \cdots (1-q^n)$ with convention $(q;q)_0$:=1.

Let $g_q(x):=1/f_q(x)$.

Question: Are there closed form expressions for $$\lim_{x \rightarrow 1} g'_q(x) \quad \mbox{and} \quad \lim_{x \rightarrow 1} g''_q(x) \quad ?$$ By a rather tricky argument, I get $\lim_{x \rightarrow 1} g'_q(x) = -(q;q)_{\infty}$. But I don't get anything simple for $\lim_{x \rightarrow 1} g''_q(x)$.

We have $$f_q(x)=(q;q)_{\infty}\sum_{n\ge 0}\left(\frac{1}{(q^{n+1};q)_{\infty}}-1\right)x^n+(q;q)_{\infty}\frac{1}{1-x}.$$ Note that $$\frac{1}{(q^{n+1};q)_{\infty}}-1=\exp\left(\sum_{\ell\ge 1}\frac{q^{(n+1)\ell}}{\ell(1-q^{\ell})}\right)-1\ll \frac{|q|^n}{(|q|;|q|)_{\infty}}$$ for sufficiently large $n$. Thus $$h_q(x):=\sum_{n\ge 0}\left(\frac{1}{(q^{n+1};q)_{\infty}}-1\right)x^n$$ is analytic for $|x|<|q|^{-1}$.
Hence for $|q|<1$, $$(q;q)_{\infty}g_q(1-x)=\frac{x}{xh_q(1-x)+1}$$ is analytic for $|1-x|<1/|q|$. Further for $|1-x|<1/|q|$, \begin{align} h_q(1-x)&=\sum_{n\ge 0}\left(\frac{1}{(q^{n+1};q)_{\infty}}-1\right)\sum_{r=0}^n(-1)^{n-r}\binom{n}{r}x^r\\ &=\sum_{r\ge 0}(-x)^r\sum_{n\ge r}(-1)^n\binom{n}{r}\left(\frac{1}{(q^{n+1};q)_{\infty}}-1\right):=\sum_{r\ge 0}\alpha_r(q)x^r. \end{align}

In particular, we have \begin{align} (q;q)_{\infty}g_q(1-x)&=\frac{x}{1+\sum_{r\ge 1}\alpha_{r-1}x^r}\\ &=x-\alpha_1(q)x^2+(\alpha_1(q)^2-\alpha_2(q))x^3+\dots \end{align} The following is easy!

• Amazing! Is this an instance of some general method, or did you invent all of it for this particular question?? – მამუკა ჯიბლაძე Jul 22 '18 at 18:18

From $q$-binomial theorem we have \begin{align} f_q(x)&=(q;q)_{\infty}\sum_{n\ge 0}\frac{x^n}{(q^{n+1};q)_{\infty}}\\ &=(q;q)_{\infty}\sum_{n\ge 0}x^n\sum_{k\ge 0}\frac{q^{(n+1)k}}{(q;q)_{k}}=(q;q)_{\infty}\sum_{k\ge 0}\frac{q^{k}}{(q;q)_{k}}\frac{1}{1-xq^k} \end{align} for all $|q|<1$ and $|x|<1$. Therefore for $|1+x|<1$ and $|x|<|q^{-k}-1|, k\ge 1$, \begin{align} \frac{f_q(1+x)}{(q;q)_{\infty}}&=-\frac{1}{x}+\sum_{k\ge 1}\frac{q^{k}}{(q;q)_{k}}\frac{1}{1-q^k-xq^k}\\ &=-\frac{1}{x}+\sum_{k\ge 1}\frac{q^{k}}{(q;q)_{k}}\frac{1}{1-q^k}\sum_{r\ge 0}\left(\frac{q^k}{1-q^k}\right)^rx^r\\ &=-\frac{1}{x}+\sum_{r\ge 0}x^r\sum_{k\ge 1}\frac{1}{(q;q)_{k}} \left(\frac{q^k}{1-q^k}\right)^{r+1}. \end{align} Now, we see that $$\frac{f_q(1+x)}{(q;q)_{\infty}}+\frac{1}{x}$$ is an analytic function for $$x\in\{z\in\mathbb{C}:|z|<|q^{-k}-1|, k=1,2,\dots\}:=\Omega_q.$$ It is clear that $\Omega_q\neq\emptyset$ is an open set for $|q|<1$. We further have \begin{align} \frac{xf_q(1+x)}{(q;q)_{\infty}}&=-1+\sum_{r\ge 1}x^{r}\sum_{k\ge 1}\frac{1}{(q;q)_{k}} \left(\frac{q^k}{1-q^k}\right)^{r}:=-1+\sum_{r\ge 1}A_r(q)x^r. \end{align} Hence $$(q;q)_{\infty}g_q(1+x)=\frac{-x}{1-\sum_{r\ge 1}A_r(q)x^r}=-x-A_1(q)x^2-(A_2(q)+A_1(q)^2)x^3-\dots$$ is an analytic function at $x=0$. Thus, $$g_q'(1)=-\frac{1}{(q;q)_{\infty}}\quad\mbox{and}\quad g_q''(1)=-\frac{2}{(q;q)_{\infty}}\sum_{k\ge 1}\frac{1}{(q;q)_{k}}\frac{q^k}{1-q^k}.$$

• Care to elaborate how you're applying the $q$-binomial theorem? I don't see any $q$-binomial coefficients around... – darij grinberg Aug 19 '18 at 11:22
• Ah, I see. Your second equality sign follows from the identity $\dfrac{1}{\left(x;q\right)_\infty} = \sum\limits_{k\geq 0} \dfrac{x^k}{\left(q;q\right)_k}$, which appears on en.wikipedia.org/wiki/Q-Pochhammer_symbol#Identities . I still don't see what this identity has to do with the $q$-binomial identity, but at least I see how to prove it: For any $m \geq 0$ and $n \geq 0$, the coefficient of $x^m q^n$ on the left hand side counts the partitions of $n$ into at most $m$ parts, while the coefficient of $x^m q^n$ on the right ... – darij grinberg Aug 19 '18 at 11:27
• ... side counts the partitions of $n$ such that each parts is $\leq m$. But there is a bijection between the former and the latter partitions, given by conjugation (= transposition). – darij grinberg Aug 19 '18 at 11:28
• From what I understand, $f_q\left(1-x\right)$ is not well-defined. But $\dfrac{f_q\left(1-x\right)}{\left(q;q\right)_\infty}$ is well-defined (as a formal power series in $q$ and $x$), because your equality $\dfrac{f_q\left(x\right)}{\left(q;q\right)_\infty} = \sum\limits_{k\geq 0} \dfrac{q^k}{\left(q;q\right)_k} \cdot \dfrac{1}{1-xq^k}$ shows that $\dfrac{f_q\left(x\right)}{\left(q;q\right)_\infty} \in \mathbb{Q}\left[x\right]\left[\left[q\right]\right]$ and we can substitute $1-x$ for the polynomial indeterminate $x$. But how do you get a $\dfrac{1}{x}$ in the result? – darij grinberg Aug 19 '18 at 11:38
• @ darij grinberg $1/x$ is from $\sum_{k\ge 0}\frac{q^k}{(q;q)_k}\frac{1}{1-xq^k}$ with $k=0$ and substitute $1−x$ to $x$. – Zhou Aug 19 '18 at 12:42