Prove - $$∑_{n=0}^{∞} \frac{(a;q)_n}{(q;q)_n} q^{n\choose 2} q^n={(−q;q)_∞}{(aq;q^2)_∞}.$$

where $(a;q)$ are the q-Pochhammer symbols.

I know that the RHS is the product of generating functions of partitions with distinct integers and distinct partitions using only odd numbers. The LHS is also recognizable as generating function of partitions. But without recognizing the summation as coefficient of a power of "x" it is difficult to recognize the LHS as any combinatorial quantity.

Can somebody give me some direction?


This identity is called Lebesgue's Identity.

Analytic Proof

Let's start with the Jackson Transformation formula ; see Gasper's book Basic Hypergeometric Series (Section 1.5) if you want to know how to prove this identity. I use the compact q-Pochhammer symbol $(a)_{n} = (a;q)_{n}$.

Jackson Transformation Formula $$\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}(q)_{n}}z^{n} = \frac{(az)_{\infty}}{(z)_{\infty}}\sum_{n=0}^{\infty}\frac{(a)_{n}(c/b)_{n}(-bz)^{n}}{(q)_{n}(c)_{n}(az)_{n}}q^{\binom{n}{2}}$$

Using this formula and the transformation $z \leftarrow -\frac{q}{b}$ and $c \leftarrow \frac{aq}{b}$ :

\begin{align} \sum_{n=0}^{\infty} \frac{(a)_{n}(aq/b^{2})_{n}}{(q)_{n}(aq/b)_{n}(-aq/b)_{n}}q^{\binom{n+1}{2}} & = \frac{(-q/b)_{\infty}}{(-aq/b)_{\infty}}\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(q)_{n}(aq/b)_{n}}\left(-\frac{q}{b}\right)^{n}\\ & = \frac{(-q/b)_{\infty}}{(-aq/b)_{\infty}}\frac{(-q)_{\infty}(aq^{2}/b^{2};q^{2})_{\infty}(aq;q^{2})_{\infty}}{(-q/b)_{\infty}(aq/b)_{\infty}} \end{align} using q-Kummer's identity (see Appendix II of 1). $$\sum_{n=0}^{\infty}\frac{(a)_{n}(aq/b^{2})_{n}}{(q)_{n}(a^{2}q^{2}/b^{2};q^{2})_{n}}q^{\binom{n+1}{2}} = \frac{(-q)_{\infty}(aq;q^{2})_{\infty}(aq^{2}/b^{2};q^{2})_{\infty}}{(a^{2}q^{2}/b^{2};q^{2})_{\infty}}$$ and, with the transformation $\frac{aq}{b^{2}} \leftarrow b$, one obtains a q-Gauss identity : $$\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(q)_{n}(abq;q^{2})_{n}}q^{\binom{n+1}{2}}=\frac{(-q)_{\infty}(aq;q^{2})_{\infty}(bq;q^{2})_{\infty}}{(abq;q^{2})_{\infty}}$$ Finally, with $b \rightarrow 0$, one gets Lebesgue's identity $$\sum_{n=0}^{\infty}\frac{(a)_{n}}{(q)_{n}}q^{\binom{n+1}{2}} = (-q)_{\infty}(aq;q^{2})_{\infty}$$

Combinatorial Proof

For a combinatorial interpretation of this identity in terms of partitions, see Tak's survey about Partition bijections (Section 4.1)


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