Stirling's Formula/Refinement/Proof 1

Theorem

A refinement of Stirling's Formula is:

$n! = \sqrt {2 \pi n} \paren {\dfrac n e}^n \paren {1 + \dfrac 1 {12 n} + \map \OO {\dfrac 1 {n^2} } }$

Proof

Let:

$\ds \map f n := \dfrac {n!} {\sqrt {2 \pi n} n^n e^{-n} } - \paren {1 + \frac 1 {12 n} }$

We need to show:

$\ds \map f n = \map \OO {\dfrac 1 {n^2} }$

Recall Limit of Error in Stirling's Formula:

$e^{1 / \paren {12 n + 1} } \le \dfrac {n!} {\sqrt {2 \pi n} n^n e^{-n} } \le e^{1 / 12 n}$

Furthermore, observe:

\(\ds e^{1 / {12 n + 1} }\)

\(\ge\)

\(\ds 1 + \frac 1 {12 n + 1}\)

Exponential Function Inequality

\(\ds \)

\(=\)

\(\ds 1 + \frac 1 {12 n} - \frac 1 {12 n \paren {12 n + 1} }\)

\(\ds \)

\(\ge\)

\(\ds \paren {1 + \frac 1 {12 n} } - \frac 1 {n^2}\)

On the other hand:

\(\ds e^{1 / {12 n} }\)

\(=\)

\(\ds 1 + \frac 1 {12 n} + \frac {e^\xi}{2!} \paren {\frac 1 {12 n} }^2\)

for a $\xi \in \closedint 0 {\frac 1 {12n} }$ by Taylor's Theorem

\(\ds \)

\(\le\)

\(\ds 1 + \frac 1 {12 n} + \frac {e^1} {2!} \paren {\frac 1 {12 n} }^2\)

Exponential is Strictly Increasing

\(\ds \)

\(\le\)

\(\ds \paren {1 + \frac 1 {12 n} } + \frac 1 {n^2}\)

$e=2.71\ldots$

Altogether we have:

$-\dfrac 1 {n^2} \le e^{1 / \paren {12 n + 1} } - \paren {1 + \dfrac 1 {12 n} } \le \map f n \le e^{1 / \paren {12 n} } - \paren {1 + \dfrac 1 {12 n} } \le \dfrac 1 {n^2}$

Therefore:

$ \size {\map f n} \le \dfrac 1 {n^2}$

$\blacksquare$

Examples

Factorial of $8$

The factorial of $8$ is given by the refinement to Stirling's formula as:

$8! \approx 40 \, 318$

which shows an error of about $0.005 \%$.

Source of Name

This entry was named for James Stirling.

Also see

• Stirling's Formula for Gamma Function