Proof of certain identities in combinatory analysis
Proceedings of the Cambridge Philosophical Society, XIX, 1919,214 – 216
Let
\begin{align}
G(x)& = 1
+\sum^\infty_1 (-1)^\nu x^{2\nu}q^{\frac{1}{2}\nu(5\nu-1)}(1-xq^{2\nu})
\frac{(1-xq)(1-xq^2)\cdots(1-xq^{\nu-1})}{(1-q)(1-q^2)(1-q^3)
\cdots(1-q^\nu)}\notag\\
&= 1- x^2q^2(1-xq^2) \frac{1}{1-q} + x^4q^9 (1-xq^4)
\frac{1-xq}{(1-q)(1-q^2)}-\cdots.
\end{align}
If we write
$$ 1 - xq^{2\nu} = 1 - q^\nu + q^\nu (1-xq^\nu),$$
every term in (1) is split up into two parts. Associating the second
part of each term with the first part of the succeeding term, we
obtain
\begin{equation}
G(x)\! =\! (1-x^2q^2)-x^2q^3 (1-x^2q^6)\frac{1-xq}{1-q}
+x^4q^{11}(1-x^2q^{10})\frac{(1-xq)(1-xq^2)}{(1-q)(1-q^2)}-\cdots.
\end{equation}
Now consider
\begin{equation}
H(x) = \frac{G(x)}{1-xq} - G(xq).
\end{equation}
Substituting for the first term from (2) and for the second term
from (1), we obtain
\begin{align*}
H(x)&= xq - \frac{x^2q^3}{1-q} \{(1-q)+xq^4 (1-xq^2)\}
+ \frac{x^4 q^{11}(1-xq^2)}{(1-q)(1-q^2)} \{(1-q^2) + xq^7 (1-xq^3)\}\\
&\quad - \frac{x^6q^{24}(1-xq^2)(1-xq^3)}{(1-q)(1-q^2)(1-q^3)} \{(1-q^3)+xq^{10}
(1-xq^4)\}+\cdots .
\end{align*}
Associating, as before, the second part of each term with the first
part of the succeeding term, we obtain
\begin{align}
H(x)& = xq (1-xq^2) \left\{1-x^2 q^6 (1-xq^4) \frac{1}{1-q}
+ x^4 q^{17} (1-xq^6) \frac{1-xq^3}{(1-q)(1-q^2)}\right.\notag\\
&\quad \left. - x^6 q^{33} (1-xq^8)
\frac{(1-xq^3)(1-xq^4)}{(1-q)(1-q^2)(1-q^3)+}\cdots
\right\}\notag\\
&= xq (1-xq^2) G(xq^2).
\end{align}
If now we write
$$ K(x) = \frac{G(x)}{(1-xq)G(xq)},$$
we obtain, from (3) and (4),
$$ K(x) = 1 + \frac{xq}{K(xq)}, $$
and so
\begin{equation}
K(x) = 1 + \frac{xq}{1+}\frac{xq^2}{1+} \frac{xq^3}{1+\cdots}.
\end{equation}
In particular we have
\begin{equation}
\frac{1}{1+}\frac{q}{1+}\frac{q^2}{1+\cdots} = \frac{1}{K(1)} =
\frac{(1-q)G(q)}{G(1)};
\end{equation}
or
\begin{equation}
\frac{1}{1+}\frac{q}{1+}\frac{q^2}{1+\cdots} =
\frac{1-q-q^4+q^7+q^{13}-\cdots}{1-q^2-q^3+q^9+q^{11}-\cdots}.
\end{equation}
This equation may also be written in the form
\begin{equation}
\frac{1}{1+}\frac{q}{1+}\frac{q^2}{1+\cdots} =
\frac{(1-q)(1-q^4)(1-q^6)(1-q^9)(1-q^{11})\cdots}
{(1-q^2)(1-q^3)(1-q^7)(1-q^8)(1-q^{12})\cdots}.
\end{equation}
If we write
$$ F(x) = \frac{G(x)}{(1-xq)(1-xq^2)(1-xq^3)\cdots},$$
then (4) becomes
$$F(x)=F(xq)+xqF(xq^2),$$
from which it readily follows that
\begin{equation}
F(x) = 1 + \frac{xq}{1-q} + \frac{x^2q^4}{(1-q)(1-q^2)} +
\frac{x^3q^9}{(1-q)(1-q^2)(1-q^3)}+\cdots .
\end{equation}
In particular we have
\begin{align}
1+ \frac{q}{1-q} + \frac{q^4}{(1-q)(1-q^2)} + \cdots & =
\frac{G(1)}{(1-q)(1-q^2)(1-q^3)+\cdots}\notag\\
&= \frac{1-q^2-q^3+q^9+q^{11}-\cdots}{(1-q)(1-q^2)(1-q^3)\cdots}\notag\\
&= \frac{1}{(1-q)(1-q^4)(1-q^6)(1-q^9)(1-q^{11})\cdots},
\end{align}
and
\begin{align}
1+ \frac{q^2}{1-q} + \frac{q^6}{(1-q)(1-q^2)} + \cdots &=
\frac{(1-q)G(q)}{(1-q)(1-q^2)(1-q^3)}\cdots\notag\\
&= \frac{1-q-q^4+q^7+q^{13}-\cdots}{(1-q)(1-q^2)(1-q^3)\cdots}\notag\\
&=\frac{1}{(1-q^2)(1-q^3)(1-q^7)(1-q^8)(1-q^{12})\cdots}.
\end{align}