# Stirling’s log-factorial approximation

In my statistical physics course, I have frequently encountered the following approximation due to Stirling:

$\ln(N!) \approx N \ln (N) -N$

It’s very useful but my professor didn’t explain how good the approximation was. The derivation I found turns out to be very simple and so I can present it in a few lines here:

1. Note that:$\int \ln(x) dx = x \ln (x) -x \quad (1)$
2. Now, if we define$S = \sum_{n=1}^{N} \ln (n) \quad (2)$
we have an upper-Riemann sum with $\Delta x =1$.
3. So we basically have the following approximation:
$S = \sum_{n=1}^{N} \ln (n \Delta x)\Delta x \approx \int_{1}^{N} \ln(x) dx \quad (3)$
4. By the intermediate value theorem,$\forall n \in\mathbb{N} \thinspace \exists c_n \in (n-1,n), S' = \sum_{n=1}^{N} \ln (c_n \Delta x)\Delta x =\int_{1}^{N} \ln(x) dx \quad(4)$where $\Delta x =1$ as defined previously.
5. Let’s check how good this approximation is:$|S - S'| = | \sum_{n=1}^{N} \ln (n) - \sum_{n=1}^{N} \ln (c_n) |$$\Rightarrow |S - S'| \leq | \sum_{n=1}^{N} \ln (n) -\ln (n+1)| = \ln (N+1) \quad (5)$
6. This error grows very slowly. In fact, if $N = 10^{24}$ i.e. the number of molecules in a glass of water,$| \ln(N!) - (N \ln (N) -N)| < 60$ which is a minuscule error relative to the number of molecules.