[Calculation 14] q-Series and the q-Binomial Theorem

In this post, we introduce q-Series and the q-Binomial theorem. For any complex number $a$, we write
$$
(a;q)_k = (1-a)(1-aq)(1-aq^2) \cdots (1-aq^{k-1})
$$
where $|q|<1$. Also we write
$$
(a;q)_\infty = \prod_{k=0}^\infty (1-aq^k).
$$
With these notations, we will prove the following theorem, which is called q-Binomial Theorem discovered by Rothe.
 

 
Proof. First, we set the function $f_a$ as follows:
$$
f_a(x) = \sum_{k=0}^\infty \frac{(a;q)_k}{(q;q)_k} x^k.
$$
Then we have
\begin{eqnarray*}
\frac{f_a(x) – f_a (qx)}{x}
&=&
\sum_{k=0}^\infty \frac{(a;q)_k}{(q;q)_k} (1-q^k) x^{k-1}\\
&=&
(1-a) \sum_{k=1}^\infty \frac{(aq;q)_{k-1}}{(q;q)_{k-1}} x^{k-1}\\
&=&
(1-a) \sum_{k=0}^\infty \frac{(aq;q)_k}{(q;q)_k} x^k\\
&=&
(1-a) f_{aq} (x)
\end{eqnarray*}
so that
\begin{equation}\tag{1}
f_a(x) – f_a (qx) = (1-a)xf_{aq} (x).
\end{equation}
Also we have
\begin{eqnarray*}
f_a (x) – f_{aq} (x)
&=&
\sum_{k=0}^\infty \frac{(aq;q)_{k-1}}{(q;q)_k} (1-a – 1+aq^k)x^k\\
&=&
-a \sum_{k=0}^\infty \frac{(aq;q)_{k-1}}{(q;q)_k} (1-q^k) x^k\\
&=&
-a \sum_{k=1}^\infty \frac{(aq;q)_{k-1}}{(q;q)_{k-1}} x^k\\
&=&
-a \sum_{k=0}^\infty \frac{(aq;q)_{k}}{(q;q)_{k}} x^{k+1}\\
&=&
-axf_{aq} (x)
\end{eqnarray*}
so that
\begin{equation}\tag{2}
f_a (x) = (1-ax) f_{aq} (x).
\end{equation}
Solving (1) and (2) to eliminate $f_{aq}(x)$, we get
$$
f_a (x) = \frac{1-ax}{1-x} f_a (qx).
$$
Iterating this relation $n$ times, we arrive at
$$
f_a (x) = \frac{(ax;q)_n}{(x;q)_n} f_a (q^n x).
$$
Taking the limit $n\to \infty$, we obtain
$$
f_a (x) = \lim_{n \to \infty} \frac{(ax;q)_n}{(x;q)_n} f_a (q^n x)
=
\frac{(ax;q)_\infty}{(x;q)_\infty},
$$
which proves the theorem.$\square$

 
References.
[1] G. E. Andrews, Special Functions, Cambridge University Press, pp. 487-489.
[2] Bruce C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, pp. 11-12.