For $|q| < 1$, $|t|<1$, then $$1+\sum_{n=1}^{\infty} \frac{(1-a)(\cdots)(1-aq^{n-1})t^n}{(1-q)(\cdots)(1-q^n)}=\prod_{n=0}^{\infty}\frac{(1-atq^n)}{(1-tq^n)}=\sum_{n=0}^{\infty}A_nt^n$$
Fromwhat I am reading they show that $A_n-A_{n-1}=q^nA_n-aq^{n-1}A_{n-1} \implies A_n = \frac{(1-aq^{n-1})}{(1-q^n)}A_{n-1}$ I cannot see how this is the case , what exactly does $\sum_{n=0}^{\infty}A_n$ equal to? I tried the following:
$$1+\sum_{n=1}^{\infty} \frac{(1-a)(\cdots)(1-aq^{n-1})t^n}{(1-q)(\cdots)(1-q^n)} = 1+\sum_{n=0}^{\infty} \frac{(1-a)(\cdots)(1-aq^{n})t^{n+1}}{(1-q)(\cdots)(1-q^{n+1})}$$ when shifting $n-1=n$, then I subtract $1$ on both sides to get $$\sum_{n=0}^{\infty}A_nt^n=\sum_{n=0}^{\infty} \frac{(1-a)(\cdots)(1-aq^{n})t^{n+1}}{(1-q)(\cdots)(1-q^{n+1})}=\sum_{n=1}^{\infty} \frac{(1-a)(\cdots)(1-aq^{n-1})t^n}{(1-q)(\cdots)(1-q^n)}$$ However this produces $t^{n+1}$ for starting from $n=0$, unlike the above equation, and isn't the RHS $A_{n-1}$?
The question is about
The equation is usually written using the q-Pochhammer symbol as
$$ F(a, t;q) := \sum_{n=0}^\infty \frac{(a;q)_n}{(q;q)_n} t^n =\prod_{n=0}^{\infty}\frac{(1-atq^n)}{(1-tq^n)}=\sum_{n=0}^{\infty}A_nt^n. \tag1 $$
Note that using the infinite product gives the equation
$$ F(a, t;q) = \frac{1-at}{1-t} F(a, qt;q). \tag2 $$
Now using the infinite sums gives equations
$$ \sum_{n=0}^\infty A_n (1-t)t^n = \sum_{n=0}^\infty A_n (1-at)(qt)^n. \tag3$$
Expanding both sides as a power series in $t$ and equating coefficients gives the equation
$$ A_n-A_{n-1}=q^nA_n-aq^{n-1}A_{n-1}. \tag4$$