Let $p(n)$ be the number of partitions of $n$, it is known that $p(n)\sim\frac{1}{4n\sqrt{3}}e^{\pi\sqrt{\frac{2n}{3}}}$, I am looking for a somewhat simple proof that $\log p(n)\sim \pi\sqrt{\frac{2n}{3}}$. Here is my first attempt :
Let $\Phi(x)=\sum_{n=0}^{+\infty}p(n)x^n=\prod_{k=1}^{+\infty}\frac{1}{1-x^k}$ for $x\in]-1,1[$, then $\forall x\in]0,1[,p(n)\leqslant\Phi(x)x^{-n}$ so we can look at how $\Phi$ behaves near $1^-$. In fact, it is not too hard to show that $\log\Phi(x)\underset{x\to 1^-}{\sim}\frac{\pi^2}{6(1-x)}$, so, if $\alpha>0$, we have $$ \log p(n)\leqslant\log\Phi\left(1-\frac{\alpha}{\sqrt{n}}\right)-n\log\left(1-\frac{\alpha}{\sqrt{n}}\right)\underset{n\to +\infty}{\sim}\left(\frac{\pi^2}{6\alpha}+\alpha\right)\sqrt{n}. $$ The $\alpha$ that minimizes the RHS is $\alpha=\frac{\pi}{\sqrt{6}}$ for which we get $\pi\sqrt{\frac{2n}{3}}$, this is suprisingly the right upper bound.
Can we obtain the right lower bound using the same ideas ? Or with some more new ideas if you want. Maybe there exists a tauberian theorem that would perfectly answer this question ?
