Stirling's Formula/Proof 2

Theorem
The factorial function can be approximated by the formula:
 * $n! \sim \sqrt {2 \pi n} \left({\dfrac n e}\right)^n$

where $\sim$ denotes asymptotically equal.

Proof
Consider the sequence $\left \langle {d_n} \right \rangle$ defined as $\displaystyle d_n = \ln \left({n!}\right) - \left({n + \frac 1 2}\right) \ln n + n$.

What we want to do is show that $\left \langle {d_n} \right \rangle$ is decreasing.

So we examine the sign of $d_n - d_{n+1}$.

Then:

Thus $f \left({x}\right) > 0$ for $\left\vert{x}\right\vert < 1$ and so:
 * $\forall n \in \N: d_n - d_{n+1} \ge 0$

and so $\left\langle{d_n}\right\rangle$ is a decreasing sequence.

Also from $(1)$:

Thus:

Thus the sequence:
 * $\left \langle {d_n - \dfrac 1 {12 n}}\right\rangle$

is increasing.

In particular:
 * $\forall n \in \N_{>0}: d_n - \dfrac 1 {12 n} \ge d_1 = \dfrac 1 {12}$

and so $\left\langle{d_n}\right\rangle$ is bounded below.

From the Monotone Convergence Theorem (Real Analysis), it follows that $\left\langle{d_n}\right\rangle$ is convergent.

Let $d_n \to d$ as $n \to \infty$.

From Exponential Function is Continuous, we have:
 * $\exp d_n \to \exp d$ as $n \to \infty$

Let $C = \exp d$.

Then:
 * $\dfrac {n!} {n^n \sqrt n e^{-n} } \to C$ as $n \to \infty$

It remains to be shown that $C = \sqrt {2 \pi}$.

Lemma 2
Finally:

Hence the result.