Deriving a less precise form of Stirling's formula: $n!=\alpha \sqrt{n}\bigg(\frac{n}{e}\bigg)^n e^\frac{\theta(n)}{12n}$

Question

I am trying to derive the less precise form of Stirling's formula asserting that there is a positive constant $\alpha$ such that $$n!=\alpha \sqrt{n}\bigg(\frac{n}{e}\bigg)^n e^\frac{\theta(n)}{12n}$$ holds for some $\theta(n)\in (0,1)$.

The hint says to deduce that $$0<\log\bigg(\frac{a_n}{a_{n+k}}\bigg)<\frac{1}{12}\bigg(\frac{1}{n}-\frac{1}{n+k}\bigg)$$ where $a_n:=n!\bigg(\frac{e}{n}\bigg)^n n^{-1/2}$ and go on to prove the existence of a positive number $\alpha$ to which the $a_n$ decrease, and finally deduce the formula.

I've deduced the inequality, but do not know how to proceed as to prove the existence of the limit for $a_n$, and how to use this fact to deduce the formula. I would greatly appreciate any help.

Answer 1

By the monotonicity of the logarithm,

$$0 = \log 1 < \log\biggl(\frac{a_n}{a_{n+k}}\biggr)$$

implies $\dfrac{a_n}{a_{n+k}} > 1$, or $a_n > a_{n+k}$, so $(a_n)$ is strictly decreasing. Further,

$$0 < \log \biggl(\frac{a_n}{a_{n+k}}\biggr) = \log a_n - \log a_{n+k} < \frac{1}{12}\bigl(\frac{1}{n} - \frac{1}{n+k}\biggr) < \frac{1}{12n}$$

shows that $(\log a_n)$ is a Cauchy sequence. Thus

$$\lambda := \lim_{n\to\infty} \log a_n$$

exists. By the continuity of the exponential function, it follows that

$$\alpha := e^{\lambda} = \exp \left(\lim_{n\to\infty} \log a_n\right) = \lim_{n\to\infty} \exp(\log a_n) = \lim_{n\to\infty} a_n.$$

Further, we have

$$0 < \log a_n - \lambda < \frac{1}{12n}\tag{1}$$

for all $n$, from which we deduce

$$\log a_n = \lambda + \frac{\theta(n)}{12n}\tag{2}$$

with

$$\theta(n) = 12n(\log a_n - \lambda) \in (0,1).$$

Applying the exponential function to $(2)$ yields

$$a_n = \alpha\cdot \exp \biggl(\frac{\theta(n)}{12n}\biggr),$$

which after rearranging becomes the desired

$$n! = \alpha \sqrt{n}\biggl(\frac{n}{e}\biggr)^n\exp\biggl(\frac{\theta(n)}{12n}\biggr).$$

We may need to still argue for the strict inequalities in $(1)$. From the monotonicity of the sequence, we have $\log a_n - \lambda > \log a_n - \log a_{n+1} > 0$, so the left inequality in $(1)$ is strict. For the strictness of the right inequality, note that

$$\log a_{n+1} - \log a_{n+k} < \frac{1}{12(n+1)}$$

for all $k \geqslant 1$ yields

$$\log a_{n+1} - \lambda \leqslant \frac{1}{12(n+1)},$$

and then we have

$$\log a_n - \lambda = (\log a_n - \log a_{n+1}) + (\log a_{n+1} - \lambda) < \frac{1}{12}\biggl(\frac{1}{n} - \frac{1}{n+1}\biggr) + \frac{1}{12(n+1)} = \frac{1}{12n}.$$