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
$$G(x)\! =\! (1-x^2q^2)-x^2q^3 (1-x^2q^6)\frac{1-xq}{1-q} +x^4q^{11}(1-x^2q^{10})\frac{(1-xq)(1-xq^2)}{(1-q)(1-q^2)}-\cdots.$$
Now consider
$$H(x) = \frac{G(x)}{1-xq} - G(xq).$$
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
$$K(x) = 1 + \frac{xq}{1+}\frac{xq^2}{1+} \frac{xq^3}{1+\cdots}.$$
In particular we have
$$\frac{1}{1+}\frac{q}{1+}\frac{q^2}{1+\cdots} = \frac{1}{K(1)} = \frac{(1-q)G(q)}{G(1)};$$
or
$$\frac{1}{1+}\frac{q}{1+}\frac{q^2}{1+\cdots} = \frac{1-q-q^4+q^7+q^{13}-\cdots}{1-q^2-q^3+q^9+q^{11}-\cdots}.$$
This equation may also be written in the form
$$\frac{1}{1+}\frac{q}{1+}\frac{q^2}{1+\cdots} = \frac{(1-q)(1-q^4)(1-q^6)(1-q^9)(1-q^{11})\cdots} {(1-q^2)(1-q^3)(1-q^7)(1-q^8)(1-q^{12})\cdots}.$$
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
$$F(x) = 1 + \frac{xq}{1-q} + \frac{x^2q^4}{(1-q)(1-q^2)} + \frac{x^3q^9}{(1-q)(1-q^2)(1-q^3)}+\cdots .$$
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}