Stirling's Factorial Approximation Formula

Stirling's Factorial Approximation Formula

In the study of combinatorics, we will see that The Factorial Function, $n!$, comes up rather frequently. As a brief reminder, recall that for $n \in \{0, 1, 2, ... \}$ we have that $n!$ is defined to be the finite product:

(1)
\begin{align} \quad n! = n \cdot (n - 1) \cdot ... \cdot 2 \cdot 1 \end{align}

It would be nice if there existed a more compact way to compute $n!$ directly for large $n$ - or at the very least, approximate $n!$ accurately. Fortunately, such a formula exists known as Stirling's Formula which is given below to be:

(2)
\begin{align} \quad n! \approx \sqrt{2\pi n}\left ( \frac{n}{e} \right )^n \end{align}

The graph of Stirling's formula is given below alongside the ordered pairs $(n, n!)$ for $n = 1, 2, 3$. We claim that Stirling's formula approximates $n!$ increasingly well for large $n$, that is:

(3)
\begin{align} \quad \lim_{n \to \infty} \frac{n!}{\sqrt{2\pi n}\left ( \frac{n}{e} \right )^n} = 1 \end{align}

We will now prove this fact as it involves some rather involved calculus. Instead, we will look at an example of applying Stirling's formula.

Suppose that we want to approximate $11!$. By using Stirling's formula we see that:

(4)
\begin{align} \quad \sqrt{22\pi} \left ( \frac{11}{e} \right )^{11} = 39615625.0506 \end{align}

Meanwhile we have that:

(5)
\begin{align} \quad 11! = 11 \cdot 10 \cdot 9 \cdot ... \cdot 2 \cdot 1 = 39916800 \end{align}

As we can see $11! \approx \sqrt{22\pi} \left ( \frac{11}{e} \right )^{11}$.