Is this $ \lim_{n\to \infty} [(n+1) f(n+1) - nf(n) ]$ reimann summable? Here, $f(n) = e^{{1 \over n} (\sum_{r=1}^{n}\ln({r\over n}))}$. So while trying to solve a limit problem I ran into this expression  $$  \lim_{n\to \infty} [(n+1) e^{{1 \over n+1} (\sum_{r=1}^{n+1}\ln({r\over n+1}))} - (n) e^{{1 \over n} (\sum_{r=1}^{n}\ln({r\over n}))}  ].$$
What I want to know is that if this is reimann summable or solvable in any way? 
It would help if you could justify your answer rigorously.
 A: Let's observe that $$f(n) \to\exp\left(\int_{0}^{1}\log x\, dx\right) =\frac{1}{e}$$ and expression under limit in question can be written as $$f(n+1)+n\{f(n+1)-f(n)\}$$ whose limit is same as that of $$\frac{1}{e}+nf(n)\frac{x_n-1}{\log x_n} \cdot\log x_n\tag{1}$$ where $$x_n=\frac{f(n+1)}{f(n)}\to 1$$ Thus the desired limit crucially depends on the limit of $$n\{\log f(n+1)-\log f(n)\} \tag{2}$$ and one can prove that the above tends to $0$ so that the desired limit is $1/e$.
We can write $(2)$ as $$\left(1-\frac{1}{n+1}\right)\sum_{k=1}^{n+1}\log\frac{k}{n+1}-\sum_{k=1}^{n}\log\frac{k}{n}$$ or $$\log\frac{(n+1)!}{(n+1)^{n+1}}-\log\frac{n!}{n^n}-\frac{1}{n+1}\sum_{k=1}^{n+1}\log\frac{k}{n+1}$$ which equals $$n\log\frac{n}{n+1}-\log f(n+1)$$ The above clearly tends to $-1-(-1)=0$ and our job is done.

On the other hand if we look closely at your $f(n) $ we see that $$nf(n) =\sqrt[n] {n!} $$ and thus your question is essentially the same as this one. Why do ask the same question twice in two seemingly different forms? At the same time I don't think my answer is fundamentally different from the answer given there by user medicu. 
A: Considering
$$a_p=p\, e^{{1 \over p} \sum_{r=1}^{p}\log\left({r\over p}\right)}$$ we have, using Pochhammer symbols 
$$\sum_{r=1}^{p}\log\left({r\over p}\right)=\log \left(\left(\frac{1}{p}\right)^p (1)_p\right)$$ which makes
$$a_p=\left((1)_p\right){}^{\frac{1}{p}}\implies \log(a_p)={\frac{1}{p}}\log\left((1)_p\right)$$ Using Stirling approximations for large $p$
$$\log\left((1)_p\right)=p (\log (p)-1)+\frac{1}{2} \log (2 \pi  p)+\frac{1}{12
   p}+O\left(\frac{1}{p^3}\right)$$
$$\log(a_p)=\log (p)-1+\frac{\log (2 \pi  p)}{2 p}+\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$
$$a_p=e^{\log(a_p)}=\frac{p}{e}+\frac{\log (2 \pi  p)}{2 e}+\frac{3 \log ^2(2 \pi  p)+2}{24 e   p}+O\left(\frac{1}{p^2}\right)$$ Apply it twice and continue with Taylor expansions
$$a_{n+1}-a_n=\frac{1}{e}+\frac{1}{2 e n}+O\left(\frac{1}{n^2}\right)$$
You could do the same with factorials since
$$\sum_{r=1}^{p}\log\left({r\over p}\right)=\sum_{r=1}^{p}\log(r)-\sum_{r=1}^{p}\log(p)=\log(p!)-p\log(p)$$
