Problems about Stirling's Series Made Easy by Chris Impens I'm reading the paper Stirling's Series Made Easy  by Chris Impens.
At the end of the proof of Theorem 1, I can only understand that $$r_n=\ln\frac{n!e^n}{\sqrt{2\pi}n^{n+\frac{1}{2}}},n=1,2,\dots$$ is strictly decreasing.
How to proof $r_n\rightarrow0$?
 A: Since $r_n$ is decreasing and bounded, $e^{r_n}$ is decreasing and bounded, thus convergent. For now, let's say that it converges to $L$ so that
$$\lim_{n \to \infty} \frac{n!e^n}{\sqrt{2 \pi} n^{n+1/2}} = \lim_{n \to \infty} \frac{n!}{\sqrt{2 \pi} n^{n+1/2}e^{-n}} = L$$
This implies that 
$$n! \sim L \sqrt{2 \pi} n^{n+1/2}e^{-n} $$

Wallis' formula is given as
$$ \prod_{n=1}^\infty \frac{2n}{2n-1} \frac{2n}{2n+1} = \frac{\pi}{2} $$
If we take the partial product, rearrange things a bit, and take the limit, we can write this as
$$\frac{\pi}{2} = \lim_{k \to \infty} \frac{1}{2k+1} \frac{2^{4k} (k!)^4}{[(2k)!]^2} $$
Since we showed above that $n! \sim L \sqrt{2 \pi} n^{n+1/2}e^{-n}$, we can plug that into the limit formulation of Wallis' formula and solve for $L$, finding that $L = 1$. 
A: Assume $\enspace\displaystyle\frac{n!e^n}{n^{n+\frac{1}{2}}}\to C\enspace$ for $\enspace n\to\infty$ .
Then with $\enspace k>0\enspace $ we get $\enspace \displaystyle\frac{(kn)!e^{kn}}{(kn)^{kn+\frac{1}{2}}}\to C\enspace $ and $\enspace \displaystyle\frac{n!^k e^{kn}}{n^{kn+\frac{k}{2}}}\to C^k$ .
It follows by division $\enspace \displaystyle\left(\frac{n!^k e^{kn}}{n^{kn+\frac{k}{2}}}\right)/\left(\frac{(kn)!e^{kn}}{(kn)^{kn+\frac{1}{2}}}\right)=\frac{n!^k k^{kn+\frac{1}{2}}}{(kn)!n^{\frac{k-1}{2}}}\to C^{k-1}$ .
Be $\enspace k:=2$ : $\enspace \displaystyle\frac{n!^2 2^{2n}}{(2n)!n^{\frac{1}{2}}}\to \frac{C}{\sqrt{2}}$ .
Because of $\enspace \displaystyle\frac{\sin(\pi x)}{\pi x}=\prod\limits_{k=1}^\infty \left(1-\left(\frac{x}{k}\right)^2\right)$ we get with $\enspace \displaystyle x=\frac{1}{2}$ 
the Wallis product $\enspace \displaystyle\frac{2}{\pi}=\prod\limits_{k=1}^\infty \left(1-\left(\frac{1}{2k}\right)^2\right)=\lim\limits_{n\to\infty}\frac{(2n)!^2 (2n+1)}{2^{4n}n!^4}$ .
It follows by multiplication 
$$(\frac{C}{\sqrt{2}})^2\frac{2}{\pi}=\lim\limits_{n\to\infty}\left(\frac{n!^2 2^{2n}}{(2n)!n^{\frac{1}{2}}}\right)^2\frac{(2n)!^2 (2n+1)}{2^{4n}n!^4}=\lim\limits_{n\to\infty}\frac{2n+1}{n}=2$$ 
and therefore $\enspace C^2=2\pi\enspace $ which means $\enspace C=\sqrt{2\pi}\enspace $ and for the problem above $\enspace r_n\to 0$ . 
A: I do not know is this answers your question; so forgive me if I am off-topic.
Let me use algebra and consider $$r_x=\log\left(\frac{x! \,e^x}{\sqrt{2\pi}\,x^{x+\frac{1}{2}}}\right)$$ The derivative is $$r'_x=-\frac{1}{2 x}-\log (x)+\psi^{(0)}(x+1)$$ where appears the digamma function. The derivative is always negative.
We could also look at the second derivative $$r''_x=\frac{1}{2 x^2}-\frac{1}{x}+\psi ^{(1)}(x+1)$$ and show that it is always positive.
Is this of any help ?
