This is new to me and I have not done any asymptotic approximation. I don't understand how they get that $\frac{n}{N}$ stays close to $\frac{2}{3}$ as N goes to infinity. Also how do they do get that $\log(1+z)\sim z$ as $z\to 0$. I know this is somehow related to the Taylor Series of $\log(1+z)$ but not how though. I would like some clarification.
Edit
I think this is what is known as a linear approximation which I have seen but really never used it. So basically $\ln(1+x)\sim f(0)+f'(0)x=\ln(1)+(\frac{1}{1+0})x=x$. So I'm guessing that's how they got that.
The taylor series for $\log(1 + z)$ for $z \to 0$ equals $ z - \dfrac{z^2}{2} + \dfrac{z^3}{3} - \dfrac{z^4}{4}$ and so on
$n \sim n^* = \dfrac{\log(\dfrac{1}{2})}{\log(1- \dfrac{1}{N})} $
for $N \to \infty \quad\ \log(1-\dfrac{1}{N})$ turns to $-\dfrac{1}{N}$ because of the taylor series for $-\dfrac{1}{N} \to 0$
Therefore $n \sim \dfrac{log(\dfrac{1}{2})}{-\dfrac{1}{N}} = N*\log(2)$
So $\dfrac{n}{N} = \log(2) = 0.69 \sim \dfrac{2}{3}$
I hope this answer makes it understandable