Is there a way to simplify the given function:
$f(n):={\sum}_{x={\lceil\frac{n}{\mathrm e}\rceil}}^{n}\frac{\ln\left(\frac{n}{x+1}\right) \left(x+1\right)}{\left(x-1\right) x}$, with $n>2$
The result can be an approximation for growing n. I thought there must be a way to get rid of the sum.
I wouldn't expect there to be a simple exact answer unless some lucky coincidence happens. But it's not so hard to approximate.
First of all, if we replace $\frac{x+1}{(x-1)x}$ with $\frac1x$ then the error is of order $1/x^2$; $\log\frac n{x+1}$ is never much bigger than 1, so the error in the whole sum is at most about $1/n^2$.
Similarly, replacing $\log\frac n{x+1}$ with $\log\frac nx$ makes a small difference for large $n$.
So $f(n)\simeq\sum_{x=\lceil\frac ne\rceil}^n\frac{\log(n/x)}x$.
Now write $x=tn$; the sum is approximately $\int_{1/e}^1\frac{-\log t}{tn}\,n\,dt=-\int_{1/e}^1\frac{\log t}t\,dt=-\frac12[(\log t)^2]_{1/e}^1=\frac12$.
(A few numerical evaluations suggest that the error is of order $1/n$, which should be easy to prove by being just a bit more precise about the approximations above.)