Q-series identities #2

Question

Prove the following

$$\frac{1}{(z;q)_{\infty}} = \sum^{\infty}_{k=0} \frac{z^k}{(q;q)_k}$$

I am looking for a proof that doesn't involve the q-binomial theorem .

where

$$(a;q)_k = \prod_{n=0}^{k-1}(1-aq^n)$$

Any help is really appreciated.

Answer 1

Let $f(z) = 1/(z;q)_{\infty} = \sum_{k = 0}^{\infty}a_{k}z^{k}$, then we can see that $(1 - z)f(z) = f(qz)$ and this means that $$\sum_{k = 0}^{\infty}a_{k}z^{k} - \sum_{k = 0}^{\infty}a_{k}z^{k + 1} = \sum_{k = 0}^{\infty}a_{k}q^{k}z^{k}$$ Comparing coefficients we get $a_{k}(1 - q^{k}) = a_{k - 1}$. Using this relation recursively we get $$a_{k} = \frac{a_{0}}{(1 - q^{k})(1 - q^{k - 1})\cdots (1 - q)} = \frac{a_{0}}{(q;q)_{k}}$$ Clearly we can see that $a_{0} = 1$ (this is obtained by putting $z = 0$ in definition of $f(z)$). Thus we get $$\frac{1}{(z;q)_{\infty}} = \sum_{k = 0}^{\infty}\frac{z^{k}}{(q;q)_{k}}$$