My question is about this proof here about Stirling's formula:
A proof I found a while ago entirely relies on creative telescoping. Since $\frac{1}{n^2}-\frac{1}{n(n+1)}=\frac{1}{n^2(n+1)}$,
$$\begin{eqnarray*} \sum_{n\geq m}\frac{1}{n^2}&=&\sum_{n\geq m}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\sum_{n\geq m}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\\&+&\frac{1}{6}\sum_{n\geq m}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)-\frac{1}{6}\sum_{n\geq m}\frac{1}{n^3(n+1)^3}\tag{1}\end{eqnarray*} $$ hence: $$ \psi'(m)=\sum_{n\geq m}\frac{1}{n^2}\leq \frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}\tag{2}$$ and in a similar fashion: $$ \psi'(m) \geq \frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}-\frac{1}{30m^5}.\tag{3}$$ Integrating twice, we have that $\log\Gamma(m)$ behaves like: $$ \log\Gamma(m)\approx\left(m-\frac{1}{2}\right)\log(m)-\color{red}{\alpha} > m+\color{blue}{\beta}+\frac{1}{12m}\tag{4}$$ where $\color{red}{\alpha=1}$ follows from $\log\Gamma(m+1)-\log\Gamma(m)=\log m$.
That gives Stirling's inequality up to a multiplicative constant.
$\color{blue}{\beta=\log\sqrt{2\pi}}$ then follows from Legendre's duplication formula and the well-known identity:
$$ \Gamma\left(\frac{1}{2}\right)=2 \int_{0}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}.\tag{5}$$
I understand how they got to $\frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}-\frac{1}{30m^5}\leq\psi′\leq\frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}$, but how do I proceed from there? Do I integrate all 3 sides, or do I approximate $\psi′\approx\frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}$ and integrate that?
And isn't one of the constants $\alpha$ or $\beta$ supposed to be multiplied by $m$ because of the double integration? I also don't quite understand how they got the values of $\alpha$ and $\beta$.
Thanks in advance.
