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According to a footnote in the famous Hardy-Ramanujan paper "Asymptotic formulae in combinatory analysis", the function $f(q)=\prod_{n=1}^{\infty}\frac{1}{1-q^n}$ vanishes like $f(re^{i\theta})=o((1-r)^{1/4})$ f(re^{i\theta})=o((1-r)^{1/4-\varepsilon})$for almost all$\theta$. No proof is given, though I can't imagine Hardy would have made a statement like this without a proof in his pocket. Edit: This isn't actually hard to guess at. By Euler's pengatonal number theorem, we have$f(q)^{-1}=\sum_{n\in \mathbf{Z}}(-1)^{n}q^{n(3n-1)/2}$, so Plancherel gives$\int_{0}^{2\pi}|f(re^{i\theta})|^{-2}d\theta=2\pi\sum_{n\in \mathbf{Z}}r^{n(3n-1)} \sim 2 \pi^{3/2}3^{-1/2}(1-r)^{-1/2}.$1 According to a footnote in the famous Hardy-Ramanujan paper "Asymptotic formulae in combinatory analysis", the function$f(q)=\prod_{n=1}^{\infty}\frac{1}{1-q^n}$vanishes like$f(re^{i\theta})=o((1-r)^{1/4})$for almost all$\theta\$. No proof is given, though I can't imagine Hardy would have made a statement like this without a proof in his pocket.