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?
Best Answer
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.