Stirling's Formula/Proof 1

Proof
Let $a_n = \dfrac {n!} {\sqrt {2 n} \paren {\frac n e}^n}$.

Part 1
It will be shown that:
 * $\lim_{n \mathop \to \infty} a_n = a$

for some constant $a$.

This will imply that:


 * $\displaystyle \lim_{n \mathop \to \infty} \frac {n!} {a \sqrt{2 n} \paren {\frac n e}^n} = 1$

By applying Power Series Expansion for $\map \ln {1 + x}$:

Let $b_n = \ln a_n$.

so the sequence $\sequence {b_n}$ is (strictly) decreasing.

From $(1)$:

Hence:

and so:


 * $b_n > b_1 - \dfrac 1 4 = \dfrac 3 4 - \dfrac {\ln 2} 2$

Thus by definition $\sequence {b_n}$ is bounded below.

By Monotone Convergence Theorem, it follows that $\sequence {b_n}$ is convergent.

Let $b$ denote its limit.

Then:
 * $\displaystyle \lim_{n \mathop \to \infty} a_n = e^{\lim_{n \mathop \to \infty} b_n} = e^b = a$

as required.

Part 2
By Wallis's Product, we have:


 * $\displaystyle \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1} = \frac \pi 2$

or equivalently:


 * $(2): \quad \displaystyle \lim_{n \mathop \to \infty} \frac {2^{4 n} \paren {n!}^4} {\paren {\paren {2 n}!}^2 \paren {2 n + 1} } = \frac \pi 2$

In Part 1 it was proved that:
 * $n! \sim a \sqrt {2 n} \paren {\dfrac n e}^n$

Substituting for $n!$ in $(2)$ yields:

Hence the result.