Stirling's Formula/Proof 1

Proof
Let $a_n = \dfrac {n!} {\sqrt{2 n} \left({\frac n e}\right)^n}$.

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

for some constant $a$.

This will imply that:


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

By applying Power Series Expansion for $\ln \left({1 + x}\right)$:

Let $b_n = \ln a_n$.

so the sequence $\left\langle{b_n}\right\rangle$ 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 $\left\langle{b_n}\right\rangle$ is bounded below.

By Monotone Convergence Theorem, it follows that $\left\langle{b_n}\right\rangle$ 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:


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

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

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

Hence the result.