I want to show the following asymptotic behavior.
Let $P(q)=\sum_{n=1}^{\infty}p(n)q^n$ be the generating function of the partitions $p(n)$ and $q=\exp(2\pi i \tau).$
I want to prove the following asymptotic behavior using the modularity of the dedekind $\eta$ function.
$\eta(\tau+1)=\eta(\tau)\exp(\frac{\pi i}{12}).$
$\eta(-\frac{1}{\tau})=\eta(\tau)\sqrt{-\tau i}.$
And we get $P(q)=\frac{q^{\frac{1}{24}}}{\eta(\tau)}.$
Prove $P(q)\sim \sqrt{-\tau i}\exp(\frac{\pi i}{12\tau})$ for $\tau \to 0$ where $\tau \in \mathbb H.$
We start with the definition $\;P(\tau) := \exp(\pi i\tau/12)/\eta(\tau)\;$ where $\tau \in \mathbb H.$ Replacing $\tau$ with $-1/\tau$ gives $\;P(-1/\tau) \exp(\frac{\pi i}{12\tau}) = 1/\eta(-1/\tau).\;$ Using modularity gives $\;1/(\eta(\tau)\sqrt{\tau/i}) = 1/\eta(-1/\tau).$ Combining the last two equations gives $\;P(-1/\tau)\sqrt{\tau/i}\exp(\frac{\pi i}{12\tau}) = 1/\eta(\tau).\;$ Substituting this in the definition of $P(\tau)$ gives $\;P(\tau) = \exp(\pi i\tau/12) P(-1/\tau)\sqrt{\tau/i}\exp(\frac{\pi i}{12\tau}).$ Since the first two factors approach $1$ as $\tau\to0$ we have our asymptotic result $\;P(q)\sim \sqrt{\tau/i}\exp(\frac{\pi i}{12\tau}).$