TITLE: Compute $\lim_{n\to\infty}\frac{\tfrac{n}{1}+\tfrac{n-1}{2}+\dots+\tfrac{2}{n-1}+\tfrac{1}{n}}{\ln(n!)}$
QUESTION [5 upvotes]: How can I compute the following limit?
$$
\lim_{n\to\infty}\frac{\dfrac{n}{1}+\dfrac{n-1}{2}+\dots+\dfrac{2}{n-1}+\dfrac{1}{n}}{\ln(n!)}
$$
I have tried lots of methods, I can't get the answer.
Although I think the limit is $0$, I don't know how to explain it. Please, if someone could help me it would be fantastic.

REPLY [5 votes]: Let $H_n:= \sum\limits_{k=1}^n \frac{1}{k}$. 
$\ln(n!)\sim n(\ln(n)-1)$ by Stirling's formula. 
As for the numerator, it is $nH_n-\frac{1}{2}-\frac{2}{3}-\cdots-\frac{n-1}{n}= nH_n-(n-1)+H_n-1$. 
There are very strong estimations for $H_n$, for example $H_n= \ln(n)+\gamma+O(\frac{1}{n})$. 
Putting all this together yields that the limit is $1$, you can even obtain a nice error term that the sequence is in fact $1+O(\frac{1}{\ln(n)})$.