# High temperature expansion of the Ising model - proof of extensivity of free energy

Consider the $$2d$$ Ising model, which has partition function $$Z = \sum_{\{S_i\}}\exp\left[J\sum_{\langle ij\rangle}S_iS_j \right],$$ where $$\langle ij\rangle$$ denotes nearest neighbours, and $$\sum_{\{S_i\}}$$ a sum over all orientations of all spins.

This model has a high-temperature expansion, which casts the free energy $$F=-\log(Z)/\beta$$ as the sum of closed graphs, $$-\beta F = N\log(2\cosh^2J) + \log\sum_{\mathcal{G}\in\text{closed}}\tilde{x}^{|\mathcal{G}|}$$ where the sum `$$\mathcal{G}\in\text{closed}$$' denotes a sum over the closed graphs on the $$2d$$ lattice, with weight $$\tilde{x}^{|\mathcal{G}|}$$. $$\tilde{x}=\tanh{J}$$ and $$|\mathcal{G}|$$ is the number of links/vertices in a particular graph.

Now, I understand how to calculate the series in $$\tilde{x}$$ using combinatoric arguments, as in this question:

How are the coefficients determined in the high temperature expansion of the 2D Ising model?

In the end, the prefactor of each term must be proportional to $$N$$ by extensivity, such that $$F\propto N$$ overall. However, calculating the prefactors, one finds for example terms which are $$\propto N^2$$ (see for example the link above). How are these two facts compatible? I am thinking that there must be some cancellation between higher-order terms, and resources which I can find on the internet state this as fact.

Can this be proved for all orders?

$$\ln\sum_{\mathcal{G}\in\text{closed}} \tilde{x}^{\vert\mathcal{G}\vert} =\ln(1+ N\epsilon^4+2N\epsilon^6+N\frac{N+9}{2}\epsilon^8 + ...)=\ln(1+x)$$
When we expand $$\ln(1+x)=x-\frac{x^2}{2} + \dots$$, we can notice for example, that this term from the linear part $$\frac{N^2}{2}\epsilon^8$$ will be cancelled by the following term in the quadratic part of the expansion $$-\frac{(N\epsilon^4)^2}{2}$$.