Stirling's Formula/Proof 2/Lemma 5

Theorem

 * $\dfrac {n!} {n^n \sqrt n e^{-n} } \to \sqrt {2 \pi}$ as $n \to \infty$

Proof
By previous work done in Stirling's Formula: Proof 2 it is noted that $\left\langle{\dfrac {n!} {n^n \sqrt n e^{-n} } }\right\rangle_{n \mathop \in \N}$ is a convergent sequence.

Let $\dfrac {n!} {n^n \sqrt n e^{-n} } \to C$ as $n \to \infty$.

Let $I_n = \displaystyle \int_0^{\frac \pi 2} \sin^n x \rd x$.

Then:

From Lemma 4, we have that:
 * $\displaystyle \lim_{n \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$

from which

Hence the result.