Suppose we have a classical Hamiltonian that can be divided into an “easy” part $H\_0$ and a “difficult” part $\Delta H$ that depends on a parameter $g$:

\begin{equation}
H = H\_0 + g \Delta H ~.
\end{equation}

The partition function can be written as 

\begin{equation}
Z=\sum\_\text{states}e^{-\beta H\_0}e^{-g\beta \Delta H} ~,
\end{equation}

because it's a classical system so $H\_0$ and $\Delta H$ commute. The partition function of the easy part,

\begin{equation}
Z\_0=\sum\_\text{states} e^{-\beta H\_0} ~,
\end{equation}

is supposed to be easy to compute.

A common way to solve this problem is to work perturbatively on $g$. We find

\begin{equation}
\begin{aligned}
Z &= \sum\_\text{states} \Big( e^{-\beta H\_0}\Big\{ 1-g\beta \Delta H +\frac{1}{2}\big(g\beta \Delta H\big)^2 +\cdots\Big\} \Big) \\
&= \sum\_\text{states} e^{-\beta H\_0} \cdot \Big( 1 - g\beta\langle \Delta H \rangle\_0 + \frac{g^2\beta^2}{2}\langle\Delta H^2\rangle\_0 + \cdots \Big)
\end{aligned}
\end{equation}

Where $\langle A \rangle\_0 \equiv \frac{1}{Z\_0} \sum\_\text{states} A \, e^{-\beta H\_0}$ is the mean value of $A$ using $H\_0$.

The problem is that $H\_0 \propto N$ because it is an extensive variable, so $e^{-\beta H\_0}\, \propto \ e^{N}$ while $\langle \Delta H \rangle\_0$ only scales as $N$. When calculating the Helmholtz energy in the thermodynamic limit,

\begin{equation}
\begin{aligned}
f&= \lim\_{N\rightarrow\infty}-\frac{kT}{N}\ln(Z) \\
&= \lim\_{N\rightarrow\infty}-\frac{kT}{N}\ln(Z\_0)-\frac{kT}{N}\ln\Big( 1 - g\beta\langle \Delta H \rangle\_0 + \frac{g^2\beta^2}{2}\langle\Delta H^2\rangle\_0 \Big) ~.
\end{aligned}
\end{equation}

The first term scales as $\frac{1}{N}\ln(e^{N})$ so it is finite, but the second term scales as $\frac{1}{N}\ln(1 + N + N^2)$, so it should vanish.

**How do we use perturbation theory in the thermodynamic limit then?**