# Newman's proof of the Asymptotic Formula for the Partition Function

I'm working on Donald J. Newman's proof that $p(n) \sim \frac{1}{4\sqrt{3}n}e^{\pi\sqrt{\frac{2n}{3}}}$, as found in Chapter II of his book Analytic Number Theory.

Here's what we have so far: the generating function for general partitions is $F(z) = \sum_{n=1}^{\infty}p(n)z^n$, which is a Taylor series of some kind. We can therefore conclude that for any $n\in\mathbb{N}$, $$p(n) = \frac{1}{2\pi i}\int_C \frac{F(z)}{z^{n+1}}dz$$ for some closed curve $C$ around the origin. That's fine and looks promising. So we need $F$ in a form that lets us perform the integration. A long-known fact is that the generating function $F$ can be written as $F(z) = \prod_{k=1}^{\infty}\frac{1}{1-z^k}$.

Next, we take logs and find after some algebra that $$\ln (F(z)) = \sum_{j=1}^{\infty} \frac{1}{j}\frac{1}{z^{-j}-1}.$$ Then letting $z=e^{-w}$ for some $w\in\mathbb{C}$ with Re$(w)$ approaching $0$ from the positive side, we get $$\ln (F(e^{-w})) = \sum_{k=1}^{\infty}\frac{1}{k}\frac{1}{e^{kw}-1}.$$ Using a Taylor expansion for $\frac{1}{e^{kw}-1}$ that converges for $w$ near $0$, we find that near $0$, $$\ln (F(e^{-w})) = \frac{\pi^2}{6w} + \frac{1}{2}\ln (1-e^{-w}) + \sum_{k=1}^{\infty}\frac{1}{k}\Bigg(\frac{1}{e^{kw}-1} - \frac{1}{kw} + \frac{e^{-kw}}{2}\Bigg) .$$

This sum, according to Newman, is essentially a Riemann sum, so taking some $h>0$, he indicates the general result that if $f$ is any real function with an integral that converges from $0$ to $\infty$, then $$\sum_{k=1}^{\infty}f(kh)h - \int_{0}^{\infty}f(x)dx < hV(f),$$ where $V(f)$ is the total variation of $f$ from $0$ to $\infty$. I'm assuming this means the supremum of $f$ minus its infimum on that interval.

Then things get messier. Newman takes $w=he^{i\theta}$ for some $\theta \in \big(-\frac{\pi}{2}, \frac{\pi}{2}\big)$, which keeps its real part positive, while $h$ keeps it small. That's fine, but then he says that we can use our Riemann sum estimate to conclude that $$\sum_{k=1}^{\infty}F(khe^{i\theta})h - \int_{0}^{\infty}F(xe^{i\theta})dx < hV_{\theta}(F),$$ where $V_{\theta}(F)$ is the variation of the argument of $F$ 'along the ray'. I don't know what that means. Is the ray the real line? Is that the argument of $F$'s image, or the argument of the variable we plug into $F$, or what? How does this inequality follow from the first? Also, isn't $F$ a complex function with a complex image, and if so how do we make sense of integrating it along the real line?

Newman goes on to say that this means $$\sum_{k=1}^{\infty}F(kw)w - \int_{0}^{\infty}F(xe^{i\theta})d(xe^{i\theta}) < w V_{\theta}(F).$$ That integral looks like it can't be right, because $xe^{i\theta}$ can't vary between $0$ and $\infty$ unless $\theta = 0$. So that's puzzling too.

There are two other things Newman mentions soon afterward that I don't fully understand: first, he says that $F$ drops off like $\frac{1}{x^2}$ as it approaches $\infty$. Is he talking about the real part of $F$ for $F$ with a real variable plugged in? If so, how does he get that asymptote? And second, he mentions the formula $V_{\theta}(F) = \int_{0}^{\infty}|F'(xe^{i\theta})|$ which, because I'm not sure about this whole ray thing, it's difficult to see how to derive.

If anyone is familiar with this work of Newman's, I'd appreciate help with these questions. (I might also add a few more as I get further in the chapter -- I hope that's not bad form.)

• The ray is the line $x\mathrm e^{\mathrm i\theta}$ from the origin to infinity at an angle $\theta$ with the real axis. – joriki Jun 28 '18 at 9:56