Stirling's Formula/Refinement

Theorem

A refinement of Stirling's Formula is:

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

where $\map \OO \cdot$ is the big-$\OO$ notation.

Proof

We have:

$\forall x \in \closedint 0 1 : \dfrac {x^2} 2 \le e^x - \paren {1 + x} \le e x^2$

since:

\(\ds \dfrac {x^2} 2\)

\(=\)

\(\ds x^2 \sum_{k \mathop \ge 0} \dfrac {0^k} {\paren {k + 2} !}\)

\(\text {(1)}: \quad\)

\(\ds \)

\(\le\)

\(\ds x^2 \sum_{k \mathop \ge 0} \dfrac {x^k} {\paren {k + 2} !}\)

\(\ds \)

\(=\)

\(\ds x^2 \sum_{k \mathop \ge 0} \dfrac {x^k} {\paren {k + 2}!}\)

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop \ge 2} \dfrac {x^k } {k !}\)

\(\ds \)

\(=\)

\(\ds e^x - \paren {1 + x}\)

Taylor Series Expansion for Exponential Function

\(\ds \)

\(=\)

\(\ds x^2 \sum_{k \mathop \ge 0} \dfrac {1^k} {\paren {k + 2}!}\)

see $\paren 1$

\(\ds \)

\(\le\)

\(\ds e x^2\)

Thus the claim follows from Limit of Error in Stirling's Formula:

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

$\blacksquare$

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.

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.5$: Permutations and Factorials: Exercise $5$