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

Question

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?

Answer 1

The higher order terms indeed cancel each other.
We consider, from the post you have linked,

$$\ln\sum_{\mathcal{G}\in\text{closed}} \tilde{x}^{\vert\mathcal{G}\vert} =\ln(1+ N\epsilon^4+2N\epsilon^6+N\frac{N+9}{2}\epsilon^8 + ...)=\ln(1+x)$$

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 $-\frac{(N\epsilon^4)^2}{2}$.

The rigorous proof that this happens to all higher order terms is not trivial (But maybe someone else can provide you with a link). But I hope that this at least convinces you that your suspicion was true.