Prove that $\prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} = \sum_{n\geq 0} \frac{(xy)^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}.$

Question

Prove that $$\prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} = \sum_{n\geq 0} \frac{(xy)^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}.$$

Here I am trying the following \begin{align*} \prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} &= \prod_{i\geq 1} \left( \sum_{n \geq 0} x^{n}y^{n(2i-1)} \right)\\ &= \sum_{m\geq 0} \sum_{n \geq 0} p(2m-1,n)x^{n}y^{2m-1}\\ &= \sum_{n\geq 0}\left( \sum_{m \geq 0} p(2m-1,n)y^{2m-1}\right)x^{n} \\ &= \sum_{n\geq 0} \frac{y^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}x^{n}\\ \end{align*} However, I am not sure if this is totally correct.

Answer 1

We obtain \begin{align*} \color{blue}{\prod_{k=1}^\infty\frac{1}{1-xy^{2k-1}}}&=\prod_{k=0}^\infty\frac{1}{1-xy^{2k+1}}\\ &=\prod_{k=0}^{\infty}\frac{1}{1-\left(xy\right)\left(y^{2}\right)^k}\\ &=\frac{1}{\left(xy;y^2\right)_{\infty}}\tag{1}\\ &=\sum_{n=0}^{\infty}\frac{1}{\left(y^2;y^2\right)_n}(xy)^n\tag{2}\\ &=\sum_{n=0}^\infty\frac{1}{\prod_{k=0}^{n-1}\left(1-\left(y^2\right)^{k+1}\right)}\left(xy\right)^n\\ &\,\,\color{blue}{=\sum_{n=0}^\infty\frac{1}{\prod_{k=1}^{n}\left(1-y^{2k}\right)}\left(xy\right)^n}\\ \end{align*}

Comment:

• In (1) we use the $q$-Pochhammer symbol \begin{align*} (a;q)_{n}&=\prod_{k=0}^{n-1}\left(1-aq^k\right)\\ (a;q)_{\infty}&=\prod_{k=0}^{\infty}\left(1-aq^k\right)\qquad\qquad |a|<1,|q|<1 \end{align*}

• In (2) we use the identity \begin{align*} \color{blue}{\frac{(at;q)_{\infty}}{(t;q)_{\infty}}=\sum_{n=0}^\infty\frac{(a;q)_n}{(q;q)_n}t^n\qquad\qquad |q|<1,|t|<1}\tag{3} \end{align*} where we set $a=0$. This is Theorem 2.1 in The Theory of Partitions by G. E. Andrews. We follow the proof there and write (3) as \begin{align*} \prod_{n=0}^\infty\frac{1-atq^n}{1-tq^n}=\sum_{n=0}^\infty\prod_{k=0}^{n-1}\frac{1-aq^k}{1-q^{k+1}}t^n \end{align*} The left-hand side is a function $F(t)$ which we want to expand as generating function \begin{align*} F(t)=\prod_{n=0}^\infty \frac{1-atq^n}{1-tq^n}=\sum_{n=0}^\infty A_n t^n\tag{4} \end{align*} in the unknown $A_n=A_n(a,q)$. Multiplying (4) with $1-t$ gives \begin{align*} (1-t)F(t)&=(1-at)\prod_{n=1}^\infty \frac{1-atq^n}{1-tq^n}=(1-at)\prod_{n=0}^\infty\frac{1-atq^{n+1}}{1-tq^{n+1}}\\ &= (1-at)F(tq)\\ &=(1-at)\sum_{n=0}^\infty A_nt^nq^n\tag{5} \end{align*} Denoting with $[t^n]$ the coefficient of $t^n$ in a series and making coefficient comparison in (5) gives \begin{align*} [t^n](1-t)F(t)&=A_n-A_{n-1}\\ &=[t^n](1-at)F(tq)\\ &=q^nA_n-aq^{n-1}A_{n-1} \end{align*} and \begin{align*} A_n(1-q^n)&=A_{n-1}\left(1-aq^{n-1}\right)\tag{6}\\ A_0&=1 \end{align*} follows. From the recurrence relation (6) we obtain for $n\geq 1$: \begin{align*} \color{blue}{A_n}&=\frac{1-aq^{n-1}}{1-q^n}A_{n-1}\\ &=\frac{\left(1-aq^{n-1}\right)\left(1-aq^{n-2}\right)}{\left(1-q^n\right)\left(1-q^{n-1}\right)}A_{n-2}\\ &=\prod_{k=0}^{n-1}\frac{1-aq^k}{1-q^{k+1}}A_0\\ &\,\,\color{blue}{=\frac{(a;q)_n}{(q;q)_n}} \end{align*} and the claim (3) follows.

Answer 2

The question asks

Prove that $$\prod_{i\geq 1}\frac{1}{1-xy^{2i-1}} = \sum_{n\geq 0} \frac{(xy)^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}.$$

As a first simplifying step, replace $\,x\,$ with $\,x/y\,$ to get

$$ \prod_{i\geq 0}\frac{1}{1-x\,y^{2i}} = \sum_{n\geq 0} \frac{x^{n}}{\prod_{i=1}^{n}\left( 1-y^{2i} \right)}. \tag{1} $$

Next, replace $\,y^2\,$ with $\,q\,$ to get

$$ \prod_{i\geq 0}\frac{1}{1-x\,q^i} = \sum_{n\geq 0} \frac{x^{n}}{\prod_{i=1}^{n}\left( 1-q^i \right)}. \tag{2} $$

The Wikipedia article Q-Pochhammer symbol section on Combinatorial interpretation states

The $q$-Pochhammer symbol is closely related to the enumerative combinatorics of partitions. The coefficient of $\,q^ma^n\,$ in $$ (a;q)_\infty^{-1} = \prod_{k=0}^\infty (1-a\,q^k)^{-1} $$ is the number of partitions of $\,m\,$ into at most $\,n\,$ parts.

Since, by conjugation of partitions, this is the same as the number of partitions of $\,m\,$ into parts of size at most $\,n\,$, by identification of generating series we obtain the identity: $$ (a;q)_\infty^{-1} = \sum_{k=0}^\infty \left( \prod_{j=1}^k \frac1{1-q^j}\right)a^k = \sum_{k=0}^\infty \frac{a^k}{(q;q)_k} $$ as in the above section.

and this last identity is exactly equation $(2)$ above except that $\,k\,$ is used instead of $\,n\,$ as the summation index, $\,a\,$ is used instead of $\,x\,$ and it is written using the $q$-Pochhammer symbol on both sides.

Here is a proof in the style of the attempt in the question. Define a generating function

$$ L:=\prod_{i\geq 1}\frac{1}{1-x\,y^{2i-1}}. \tag{3} $$

Use the Taylor series of $\,\frac1{1-z}\,$ to get

$$ L= \prod_{i\geq 1} \left( \sum_{n \geq 0} x^{n}y^{n(2i-1)} \right). \tag{4} $$

Define $\,q(m,n)\,$ to be the number of partitions of $\,m\,$ into exactly $\,n\,$ odd parts to get

$$ L = \sum_{m\geq 0} \sum_{n \geq 0} q(m,n)\,x^{n}y^m. \tag{5} $$

Reverse the order of summation to get

$$ L = \sum_{n\geq 0}\left( \sum_{m \geq 0} q(m,n)\,y^m\right)x^{n}. \tag{6} $$

Use the fact that the number of partitions enumerated by $\,q(m,n)\,$ is the same as the number of partitions of $\,m-n\,$ into at most $\,n\,$ even parts, since if we subtract $1$ from each part the odd parts become even parts and the $1$ parts become $0$ parts, to get

$$ L=\sum_{n\geq 0} \frac{y^{n}}{\prod_{i=1}^{n}\left(1-y^{2i} \right)}x^{n}. \tag{7} $$

Combine the $n$th powers to get the final result

$$ L = \sum_{n\geq 0} \frac{(x\,y)^{n}}{\prod_{i=1}^{n}\left(1-y^{2i} \right)}. \tag{8} $$

Answer 3

One perspective is to simply write

$$p=\dfrac{1}{(1-xq)(1-xq^2)(1-xq^3)\cdots}=A_0+\dfrac{A_1}{1-xq}+\dfrac{A_2}{(1-xq)(1-xq^2)}+\cdots$$ and multiply each consecutive term while setting the numerator equal to $1$. This process, assuming $A_0=1$, gives

$$A_0(1-xq)+A_1=1\implies A_1=xq;$$ and that $$p=\dfrac{1}{(1-xq)\cdots(1-xq^n)}+ \dfrac{A_{n+1}}{(1-xq)\cdots(1-xq^{n+1})}+\cdots$$ for all integers $n\geq 1$, and therefore that $$A_n=xq^n.$$ While not a proof, this is a simple outline that can be easily made into one. Either this, or you can use the fact that

$$a_0\prod_{n=1}^{\infty}\left(1+\dfrac{a_n}{\sum_0^{n-1}a_k}\right)=\sum_{n=0}^{\infty}a_n,$$and put $a_n=\dfrac{xq^n}{(1-xq)\cdots (1-xq^n)}.$