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 1

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}$

$\blacksquare$

Proof 2

A refinement of Stirling's Formula is:

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

Let $z\in \R_{>0}$ and $n \in \N_{\ge 0}$

Let $\ds c_n = \ln \map \Gamma {z + n}$

We begin by observing:

$\ds \map \Gamma {z + n}$

$=$

$\ds \map \Gamma {z + 1} \times \paren {z + 1} \times \paren {z + 2} \times \cdots \times \paren {z + n - 1}$

Gamma Difference Equation

$\ds \leadsto \ \ $

$\ds \ln \map \Gamma {z + n}$

$=$

$\ds \ln \map \Gamma {z + 1} + \map \ln {z + 1} + \map \ln {z + 2} + \cdots + \map \ln {z + n - 1}$

Sum of Logarithms

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

$\ds \leadsto \ \ $

$\ds \ln \map \Gamma {z + n}$

$=$

$\ds \ln \map \Gamma {z + 1} + \sum_{k \mathop = 1}^{n - 1} \map \ln {z + k}$

Then from $\paren 1$ above, we have:

$\ds c_n = \ln \map \Gamma {z + 1} + \sum_{k \mathop = 1}^{n - 1} \map \ln {z + k}$

Therefore:

$\ds c_{n + 1} - c_n$

$=$

$\ds \paren {\ln \map \Gamma {z + 1} + \sum_{k \mathop = 1}^{\paren {n + 1} - 1} \map \ln {z + k} } - \paren {\ln \map \Gamma {z + 1} + \sum_{k \mathop = 1}^{n - 1} \map \ln {z + k} }$

$\text {(2)}: \quad$

$\ds \leadsto \ \ $

$\ds c_{n + 1} - c_n$

$=$

$\ds \map \ln {z + n}$

By the analogy between the derivative and the finite difference, we consider $c_n$ to be approximately $\ds \int \map \ln {z + n} \rd z$.

With this in mind, based upon Primitive of Logarithm of x, we set:

$\ds c_n = \paren {z + n} \map \ln {z + n} - \paren {z + n} + d_n$

Quick recap:

$\ds c_n$

$=$

$\ds \ln \map \Gamma {z + n}$

$\ds $

$=$

$\ds \ln \map \Gamma {z + 1} + \sum_{k \mathop = 1}^{n - 1} \map \ln {z + k}$

$\ds \leadsto \ \ $

$\ds c_n$

$\approx$

$\ds \int \map \ln {z + n} \rd z$

by the analogy between the derivative and the finite difference

$\text {(3)}: \quad$

$\ds \leadsto \ \ $

$\ds c_n$

$=$

$\ds \paren {z + n} \map \ln {z + n} - \paren {z + n} + d_n$

based upon Primitive of Logarithm of x

Therefore:

$\ds c_{n + 1} - c_n$

$=$

$\ds \paren {\paren {z + n + 1} \map \ln {z + n + 1} - \paren {z + n + 1} + d_{n + 1} } - \paren {\paren {z + n} \map \ln {z + n} - \paren {z + n} + d_n }$

from $\paren 3$

$\ds \leadsto \ \ $

$\ds \map \ln {z + n}$

$=$

$\ds \paren {\paren {z + n + 1} \map \ln {z + n + 1} - \paren {z + n + 1} + d_{n + 1} } - \paren {\paren {z + n} \map \ln {z + n} - \paren {z + n} + d_n }$

from $\paren 2$

$\ds $

$=$

$\ds \paren {z + n + 1} \map \ln {z + n + 1} - \paren {z + n} \map \ln {z + n} + d_{n + 1} - d_n - 1$

$\ds \leadsto \ \ $

$\ds d_{n + 1} - d_n$

$=$

$\ds 1 + \map \ln {z + n} - \paren {\paren {z + n + 1} \map \ln {z + n + 1} - \paren {z + n} \map \ln {z + n} }$

rearranging and moving $d_{n + 1} - d_n$ to the left hand side

$\ds $

$=$

$\ds 1 - \paren {\paren {z + n + 1} \map \ln {z + n + 1} - \paren {z + n + 1} \map \ln {z + n} }$

$\ds $

$=$

$\ds 1 - \paren {z + n + 1} \map \ln {\dfrac {z + n + 1} {z + n} }$

Sum of Logarithms

$\ds $

$=$

$\ds 1 - \paren {z + n + 1} \map \ln {1 + \dfrac 1 {z + n} }$

$\ds $

$=$

$\ds 1 - \paren {z + n + 1} \paren {\paren {\dfrac 1 {z + n} } - \dfrac {\paren {\dfrac 1 {z + n} }^2 } 2 + \dfrac {\paren {\dfrac 1 {z + n} }^3 } 3 - \cdots }$

Power Series Expansion for Logarithm of 1 + x

$\ds $

$=$

$\ds 1 - \paren {z + n} \paren {\paren {\dfrac 1 {z + n} } - \dfrac {\paren {\dfrac 1 {z + n} }^2 } 2 + \dfrac {\paren {\dfrac 1 {z + n} }^3 } 3 - \cdots } - \paren {\paren {\dfrac 1 {z + n} } - \dfrac {\paren {\dfrac 1 {z + n} }^2 } 2 + \dfrac {\paren {\dfrac 1 {z + n} }^3 } 3 - \cdots }$

$\ds $

$=$

$\ds 1 - \paren {\paren {1 } - \dfrac {\paren {\dfrac 1 {z + n} } } 2 + \dfrac {\paren {\dfrac 1 {z + n} }^2 } 3 - \cdots } - \paren {\paren {\dfrac 1 {z + n} } - \dfrac {\paren {\dfrac 1 {z + n} }^2 } 2 + \dfrac {\paren {\dfrac 1 {z + n} }^3 } 3 - \cdots }$

distributing $\paren {z + n}$

$\ds $

$=$

$\ds \paren {\dfrac {\paren {\dfrac 1 {z + n} } } 2 - \dfrac {\paren {\dfrac 1 {z + n} }^2 } 3 - \cdots } - \paren {\paren {\dfrac 1 {z + n} } - \dfrac {\paren {\dfrac 1 {z + n} }^2 } 2 + \dfrac {\paren {\dfrac 1 {z + n} }^3 } 3 - \cdots }$

$\ds $

$=$

$\ds \paren {\paren {\frac 1 2 - 1} \dfrac 1 {z + n} + \paren {-\frac 1 3 + \frac 1 2} \dfrac 1 {\paren {z + n}^2 } + \cdots}$

$\text {(4)}: \quad$

$\ds \leadsto \ \ $

$\ds d_{n + 1} - d_n$

$=$

$\ds \paren {-\dfrac 1 {2 \paren {z + n} } + \dfrac 1 {6 \paren {z + n}^2 } - \cdots}$

Let $\ds d_n = e_n - \dfrac 1 2 \map \ln {z + n}$.

Then:

$\ds d_{n + 1} - d_n$

$=$

$\ds \paren {e_{n + 1} - \frac 1 2 \map \ln {z + n + 1} } - \paren {e_n - \frac 1 2 \map \ln {z + n} }$

$\ds $

$=$

$\ds e_{n + 1} - e_n - \frac 1 2 \map \ln {\dfrac {z + n + 1} {z + n} }$

Sum of Logarithms

$\ds $

$=$

$\ds e_{n + 1} - e_n - \frac 1 2 \map \ln {1 + \dfrac 1 {z + n} }$

$\ds \leadsto \ \ $

$\ds e_{n + 1} - e_n$

$=$

$\ds d_{n + 1} - d_n + \frac 1 2 \map \ln {1 + \frac 1 {z + n} }$

rearranging and moving $e_{n + 1} - e_n$ to the left hand side

$\ds $

$=$

$\ds \paren {-\dfrac 1 {2 \paren {z + n} } + \dfrac 1 {6 \paren {z + n}^2} - \cdots} + \frac 1 2 \paren {\paren {\dfrac 1 {z + n} } - \dfrac {\paren {\dfrac 1 {z + n} }^2 } 2 + \cdots}$

from $\paren 4$ above and Power Series Expansion for Logarithm of 1 + x

$\ds $

$=$

$\ds \paren { \paren {\frac 1 6 - \frac 1 4 } \dfrac 1 {\paren {z + n}^2} + \OO \paren {\frac 1 {\paren {z + n}^3} } }$

where $\OO$ is big-$\OO$ notation

$\text {(5)}: \quad$

$\ds \leadsto \ \ $

$\ds e_{n + 1} - e_n$

$=$

$\ds \paren {-\dfrac 1 {12 \paren {z + n}^2} + \OO \paren {\frac 1 {\paren {z + n}^3} } }$

We now observe that:

$\ds e_n - e_0$

$=$

$\ds \paren {e_1 - e_0} + \paren {e_2 - e_1} + \paren {e_3 - e_2} + \cdots + \paren {e_{n - 1} - e_{n - 2} } + \paren {e_n - e_{n - 1} }$

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^{n - 1} \paren {e_{k + 1} - e_k }$

Definition of Telescoping Series

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^{n - 1} \paren {-\dfrac 1 {12 \paren {z + k}^2 } + \OO \paren {\frac 1 {\paren {z + k}^3} } }$

from $\paren 5$ above

Let $\ds \map {K_1} z = \lim_{n \mathop \to \infty} \paren {e_n - e_0}$.

Then:

$\ds e_n - e_0$

$=$

$\ds \sum_{k \mathop = 0}^{n - 1} \paren {e_{k + 1} - e_k}$

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^\infty \paren {e_{k + 1} - e_k} - \sum_{k \mathop = n}^\infty \paren {e_{k + 1} - e_k}$

Indexed Summation over Adjacent Intervals

$\ds \leadsto \ \ $

$\ds e_n$

$=$

$\ds e_0 + \map {K_1} z - \sum_{k \mathop = n}^\infty \paren {e_{k + 1} - e_k}$

adding $e_0$ to both sides and $\ds \map {K_1} z = \lim_{n \mathop \to \infty} \paren {e_n - e_0}$

$\ds $

$=$

$\ds e_0 + \map {K_1} z - \sum_{k \mathop = n}^\infty \paren { -\dfrac 1 {12 \paren {z + k}^2 } + \OO \paren {\frac 1 {\paren {z + k}^3} } }$

from $\paren 5$ above

$\ds $

$=$

$\ds e_0 + \map {K_1} z - \int_n^\infty \paren { -\dfrac 1 {12 \paren {z + k}^2} + \OO \paren {\frac 1 {\paren {z + k}^3} } } \rd k$

approximating the sum with an integral

$\ds $

$=$

$\ds e_0 + \map {K_1} z - \intlimits {\dfrac 1 {12 \paren {z + k} } + \OO \paren {\frac 1 {\paren {z + k}^2} } } n \infty$

Primitive of Power

$\ds $

$=$

$\ds e_0 + \map {K_1} z + \dfrac 1 {12 \paren {z + n} } + \OO \paren {\frac 1 {\paren {z + n}^2} }$

$\text {(6)}: \quad$

$\ds \leadsto \ \ $

$\ds e_n$

$=$

$\ds \map \ln {\map C z} + \dfrac 1 {12 \paren {z + n} } + \OO \paren {\frac 1 {\paren {z + n}^2} }$

let $\map \ln {\map C z} = e_0 + \map {K_1} z$

Therefore, we have:

$\ds c_n$

$=$

$\ds \paren {z + n} \map \ln {z + n} - \paren {z + n} + d_n$

from $\paren 3$ above

$\ds $

$=$

$\ds \paren {z + n} \map \ln {z + n} - \paren {z + n} + \paren {e_n - \frac 1 2 \map \ln {z + n} }$

substituting $\ds d_n = e_n - \dfrac 1 2 \map \ln {z + n}$

$\ds $

$=$

$\ds \paren {z + n - \frac 1 2} \map \ln {z + n} - \paren {z + n} + \paren {\map \ln {\map C z} + \dfrac 1 {12 \paren {z + n} } + \OO \paren {\frac 1 {\paren {z + n}^2 } } }$

from $\paren 6$ above

Exponentiating both sides, we obtain:

$\ds \map \Gamma {z + n}$

$=$

$\ds \map \exp {\paren {z + n - \frac 1 2} \map \ln {z + n} - \paren {z + n} + \paren {\map \ln {\map C z} + \frac 1 {12 \paren {z + n} } + \OO \paren {\frac 1 {\paren {z + n}^2 } } } }$

Exponential of Natural Logarithm

$\ds $

$=$

$\ds \map C z \paren {z + n}^{z + n - \frac 1 2} \map \exp {-\paren {z + n} } \map \exp {\frac 1 {12 \paren {z + n} } + \OO \paren {\frac 1 {\paren {z + n}^2 } } }$

Product of Powers

$\text {(7)}: \quad$

$\ds \leadsto \ \ $

$\ds \map \Gamma {z + n}$

$=$

$\ds \map C z \paren {z + n}^{z + n - \frac 1 2} \map \exp {-\paren {z + n} } \paren {1 + \frac 1 {12 \paren {z + n} } + \OO \paren {\frac 1 {\paren {z + n}^2 } } }$

Power Series Expansion for Exponential Function

We claim $\map C z$ is independent of $z$.

$\ds 1$

$=$

$\ds \lim_{n \mathop \to \infty} n^{-z} \frac {\map \Gamma {n + z} } {\map \Gamma n}$

$\ds $

$=$

$\ds \lim_{n \mathop \to \infty} n^{-z} \frac {\map C z \paren {z + n}^{z + n - \frac 1 2} \map \exp {-\paren {z + n} } } {\map C 0 n^{n - \frac 1 2} \map \exp {-n} }$

$\ds $

$=$

$\ds \frac {\map C z} {\map C 0} \lim_{n \mathop \to \infty} \frac { \paren {z + n}^{z + n - \frac 1 2} \map \exp {-\paren {z + n} } } { n^{z + n - \frac 1 2} \map \exp {-n} }$

$\ds $

$=$

$\ds \frac {\map C z} {\map C 0} \lim_{n \mathop \to \infty} \paren {1 + \frac z n}^{z + n - \frac 1 2} e^{-z}$

$\ds $

$=$

$\ds \frac {\map C z} {\map C 0} e^z e^{-z}$

Real Exponential Function as Limit of Sequence

$\ds $

$=$

$\ds \frac {\map C z} {\map C 0 }$

Therefore, $\map C z$ is a constant, and:

$\map \Gamma z \sim C z^{z - \frac 1 2} e^{-z}$ as $z \to \infty$

Therefore:

$\ds \map \Gamma z$

$\sim$

$\ds C z^{z - \frac 1 2} e^{-z}$

$\ds \leadsto \ \ $

$\ds z \map \Gamma z$

$\sim$

$\ds z C z^{z - \frac 1 2} e^{-z}$

multiplying both sides by z

$\ds \leadsto \ \ $

$\ds \map \Gamma {z + 1}$

$\sim$

$\ds C z^{z + \frac 1 2} e^{-z}$

Gamma Difference Equation

$\ds \leadsto \ \ $

$\ds z!$

$\sim$

$\ds C z^{z + \frac 1 2} e^{-z}$

Gamma Function Extends Factorial

To determine $C$, we use Wallis's Product:

$\ds \pi$

$=$

$\ds 2 \prod_{k \mathop = 1}^\infty \frac {2 k} {2 k - 1} \cdot \frac {2 k} {2 k + 1}$

$\ds $

$=$

$\ds 2 \prod \paren {\frac 2 1 \cdot \frac 2 3} \paren {\frac 4 3 \cdot \frac 4 5} \paren {\frac 6 5 \cdot \frac 6 7} \cdots$

$\ds $

$=$

$\ds \lim_{n \mathop \to \infty} \dfrac 2 {2 n + 1} \prod_{k \mathop = 1}^n \frac {\paren {2 k}^2} {\paren {2 k - 1}^2}$

$\ds $

$=$

$\ds \lim_{n \mathop \to \infty} \dfrac 1 {n + \frac 1 2} \prod_{k \mathop = 1}^n \frac {\paren {2 k}^2} {\paren {2 k - 1}^2} \dfrac {\paren {2 k}^2} {\paren {2 k}^2}$

multiplying by $1$

$\ds $

$=$

$\ds \lim_{n \mathop \to \infty} \dfrac 1 {n + \frac 1 2} \prod_{k \mathop = 1}^n \frac {2^4 k^4} {\paren {\paren {2 k - 1} \paren {2 k} }^2}$

$\ds \leadsto \ \ $

$\ds \pi$

$=$

$\ds \lim_{n \mathop \to \infty} \frac {2^{4 n} \paren {n!}^4} {\paren {\paren {2 n}!}^2 \paren {n + \frac 1 2} }$

$\ds \leadsto \ \ $

$\ds \sqrt {\pi}$

$=$

$\ds \lim_{n \mathop \to \infty} \sqrt {\frac {2^{4 n} \paren {n!}^4} {\paren {\paren {2 n}!}^2 \paren {n + \frac 1 2} } }$

taking the Square Root of both sides

$\text {(8)}: \quad$

$\ds \leadsto \ \ $

$\ds \sqrt {\pi}$

$=$

$\ds \lim_{n \mathop \to \infty} \frac {2^{2 n} \paren {n!}^2} {\paren {2 n}! \sqrt {n + \frac 1 2} }$

Substituting for $n!$ in $(8)$ yields:

$\ds \sqrt {\pi}$

$=$

$\ds \lim_{n \mathop \to \infty} \frac {2^{2 n} \paren {C n^{n + \frac 1 2} e^{-n} }^2} {C \paren {2 n}^{2 n + \frac 1 2} e^{- 2 n} \sqrt {n + \frac 1 2} }$

$n! = C n^{n + \frac 1 2} e^{-n}$

$\ds \leadsto \ \ $

$\ds \sqrt {\pi}$

$=$

$\ds C \lim_{n \mathop \to \infty} \frac {2^{2 n} n^{2 n + 1} e^{-2 n} } {\paren {2 n}^{2 n + \frac 1 2} e^{- 2 n} \sqrt {n + \frac 1 2} }$

Power of Product

$\ds \leadsto \ \ $

$\ds \sqrt {\pi}$

$=$

$\ds C \lim_{n \mathop \to \infty} \frac n {\paren {2 n}^{\frac 1 2} \sqrt {n + \frac 1 2} }$

$\ds \leadsto \ \ $

$\ds C$

$=$

$\ds \sqrt {2 \pi}$

Putting all of the pieces together, we have:

$\ds \map \Gamma {z + n}$

$=$

$\ds \map C z \paren {z + n}^{z + n - \frac 1 2} \map \exp {-\paren {z + n} } \paren {1 + \frac 1 {12 \paren {z + n} } + \OO \paren {\frac 1 {\paren {z + n}^2 } } }$

from $\paren 7$ above

$\ds \leadsto \ \ $

$\ds \map \Gamma {z + n}$

$=$

$\ds \sqrt {2 \pi} \paren {z + n}^{z + n - \frac 1 2} \map \exp {-\paren {z + n} } \paren {1 + \frac 1 {12 \paren {z + n} } + \OO \paren {\frac 1 {\paren {z + n}^2 } } }$

$C = \sqrt {2 \pi}$

$\ds \leadsto \ \ $

$\ds \map \Gamma z$

$=$

$\ds \sqrt {2 \pi} z^{z - \frac 1 2} e^{-z} \paren {1 + \frac 1 {12 z } + \OO \paren {\frac 1 {z^2 } } }$

setting $n = 0$

$\ds \leadsto \ \ $

$\ds z \map \Gamma z$

$=$

$\ds z \sqrt {2 \pi} z^{z - \frac 1 2} e^{-z} \paren {1 + \frac 1 {12 z } + \OO \paren {\frac 1 {z^2 } } }$

multiplying both sides by z

$\ds \leadsto \ \ $

$\ds \map \Gamma {z + 1}$

$=$

$\ds \sqrt {2 \pi} z^{z + \frac 1 2} e^{-z} \paren {1 + \frac 1 {12 z } + \OO \paren {\frac 1 {z^2 } } }$

Gamma Difference Equation

$\ds \leadsto \ \ $

$\ds z!$

$=$

$\ds \sqrt {2 \pi z} \paren {\dfrac z e}^z \paren {1 + \dfrac 1 {12 z} + \map \OO {\dfrac 1 {z^2} } }$

Gamma Function Extends Factorial

$\blacksquare$

Examples

Factorial of $8$

The factorial of $8$ is given by the as:

$8! \approx 40 \, 318$

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

Also see

Stirling's Formula for Gamma Function

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 $5$