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


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.
