q-series identities

Question

Here is my first question in this site

Prove the following

$$\lim_{q \to 1 }\frac{(a)_{\infty}}{(aq^x)_{\infty}}=(1-a)^x$$

If $x$ was an integer , then this is an easy task.

Answer 1

Ok , I think I got it , this is a simple consequence of the q-binomial theorem

Consider the following

$${}_1\phi_0 (a;- ;q,z) = \sum_{k\geq 0}\frac{(a;q)_k}{(q;q)_k}z^k=\frac{(az;q)_{\infty}}{(z;q)_{\infty}}$$ (1)

In (1) let $a = q^{x}$ and $z = a$

$${}_1\phi_0 (q^x;- ;q,a) = \sum_{k\geq 0}\frac{(q^x;q)_k}{(q;q)_k}a^k=\frac{(aq^{x};q)_ {\infty} }{(a;q)_{\infty}}$$

Hence we have

$$\frac{(aq^{x};q)_{\infty}}{(a;q)_{\infty}}= \sum_{k\geq 0}\frac{(q^x;q)_k}{(q;q)_k}a^k$$

Now consider the limit

$$ \lim_{q \to 1}\frac{(aq^{x};q)_{\infty}}{(a;q)_{\infty}}= \lim_{q \to 1} \sum_{k\geq 0}\frac{(q^x;q)_k}{(q;q)_k}a^k$$(2)

Suppose that $|a|<1$ and $|q|<1$ so the sum is uniformly convergent on any sub-disk . So we have to approach $1$ from the left to stay in the disk !

The idea is use the L'Hospitale rule

$$\lim_{q \to 1^-}\frac{(q^x;q)_k}{(q;q)_k} = \lim_{q \to 1^-} \frac{(1-q^x)\cdot (1-q^{x+1}) \cdot(1-q^{x+2}) \cdots (1-q^{x+n-1}) }{(1-q)\cdot(1-q^2)\cdot(1-q^3) \cdots(1-q^n)}$$

which can be written as

$$\lim_{q \to 1^-}\frac{(q^x;q)_k}{(q;q)_k} = \lim_{q \to 1^-} \frac{(1-q^x)}{1-q}\cdot \lim_{q \to 1^-}\frac{(1-q^{x+1})}{1-q^2} \cdot \lim_{q \to 1} \frac{(1-q^{x+2})}{1-q^3} \cdots \lim_{q \to 1^-} \frac{(1-q^{x+n-1}) }{(1-q^n)}$$

$$ \lim_{q \to 1^-}\frac{(q^x;q)_k}{(q;q)_k} = \frac{x (x+1)(x+2)\cdots (x+n-1)}{1\cdot 2 \cdot 3 \cdots n} = \frac{(x)_n}{n!} $$

Substitute in (2)

$$ \lim_{q \to 1^-}\frac{(aq^{x};q)_{\infty}}{(a;q)_{\infty}}= \sum_{k\geq 0}\frac{(x)_n}{n!}a^k$$

The sum on the right is well-know $(1-x)^{-a}$

$$ \lim_{q \to 1^-}\frac{(aq^{x};q)_{\infty}}{(a;q)_{\infty}}= (1-x)^{-a} $$ (3)

From (3) we conclude that

$$ \lim_{q \to 1^-}\frac{(a;q)_{\infty}}{ (aq^{x};q)_{\infty}}= (1-x)^{a} $$