Background
We start with the identity:
$$ \sum_{r=1}^n r \ln r + \ln (r-1)! = n\ln n! $$
$$ \implies \sum_{r=1}^n \frac{r}{n} \ln r + \frac{1}{n} \ln\Gamma(r) = \ln n!$$
$$ \implies \sum_{r=1}^n \frac{r}{n} \ln (\frac{r}{n}) + \frac{r}{n} \ln(n) + \frac{1}{n} \ln\Gamma(r) = \ln n! $$
$$ \implies \sum_{r=1}^n \frac{r}{n} \ln (\frac{r}{n}) + \frac{1}{n} \ln\Gamma(r) = -\frac{(n+1)}{2} \ln(n) + \ln n! $$
$$ \implies \sum_{r=1}^n (\frac{r}{n} \ln (\frac{r}{n}) + \frac{1}{n} \ln\Gamma(r))\frac{1}{n} = -\frac{(n+1)}{2n} \ln(n) + \frac{\ln n!}{n} $$
$$ \implies \sum_{r=1}^n (\frac{r}{n} \ln (\frac{r}{n})) \frac{1}{n} = -\frac{(n+1)}{2n} \ln(n) -\sum_{r=1}^n \frac{1}{n^2} \ln\Gamma(r) + \frac{\ln n!}{n} $$
Defining $ F(n) = \int x \ln (x) dx = \frac{n^2}{2}\ln(n) - \frac{n^2}{4} $
$$ \implies \frac{1}{F(n)}\sum_{r=1}^n (\frac{r}{n} \ln (\frac{r}{n})) \frac{1}{n} = \frac{1}{F(n)}(-\frac{(n+1)}{2n} \ln(n) + \frac{\ln n!}{n} -\sum_{r=1}^n \frac{1}{n^2} \ln\Gamma(r))) $$
Taking $\lim n \to \infty$ and defining $G(n)$ as the BarnesG function:
$$ \implies \frac{n^2}{2}\ln(n) - \frac{n^2}{4} \sim \frac{-1}{2} \ln(n) -\frac{1}{n^2} \ln G(n+2) + \frac{\ln n!}{n} $$
$$ \implies - n \ln n! + \frac{n^4}{2}\ln(n) - \frac{n^4}{4} + \frac{n^2}{2} \ln(n) \sim - \ln G(n+2) $$
$$ \implies n \ln n! - \frac{n^4}{2}\ln(n) + \frac{n^4}{4} - \frac{n^2}{2} \ln(n) \sim \ln G(n+2) $$
Question(Edited)
I found my mistake .... Is this algebric identity known? Also, is weather this relation between the factorial function and the BarnesG function is known?