I am trying to approximate the expression $2^{n^2} (2^{-n};2)_n$, where $(2^{-n};2)_n$ the q-Pochhammer symbol. Actually, I need a formula for the asymptotics of the Pochhammer symbol and $2^{n^2}$ as $n \to \infty$. One such formula is provided about halfway down Wolfram's page:
$$(x)_n \sim \frac{2\pi}{\Gamma(x)} e^{-n}n^{x+n-\frac12}(1+O(\frac1n)) \qquad n \to \infty$$
but I am not sure if it is working for me. Any help?
By using $$ (a \, q^{-n}; q)_{n} = \left(\frac{a}{q}; q\right)_{n} \, \left(\frac{-a}{q}\right)^{n} \, q^{- \binom{n}{2}}$$ then $$q^{n^{2}} \, (q^{-n}; q)_{n} = (-1)^n \, q^{\binom{n}{2}} \, (q; q)_{n} = q^{\binom{n}{2}} \, \prod_{j=1}^{n} (q^j -1)$$ This is, in essence, $$q^{n^{2}} \, (q^{-n}; q)_{n} \geq q^{\binom{n}{2}} \, q^{n-1} = q^{\binom{n+1}{2} -1}$$ and for the case of $q=2$ then $$2^{n^2} \, (2^{-n}; 2)_{n} \geq 2^{\binom{n+1}{2} -1}.$$
