Compute $\lim_{n\to\infty}\frac{\tfrac{n}{1}+\tfrac{n-1}{2}+\dots+\tfrac{2}{n-1}+\tfrac{1}{n}}{\ln(n!)}$

Question

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.

Can you take a screenshot of what you tried and upload it to us? We like to see what your efforts have been so that we can better help you with this problem. — Noble Mushtak, Dec 28 '18 at 21:04
For what it's worth, Wolfram Alpha says the sum on the top is $(n+1)H_n-n$. $H_n\approx \ln n$ for big n, so this gives us $n\ln n+\ln n-n$ over $\ln(n!)$. $\ln(n!)$ is about $n\ln n$, so I'm guessing the limit is $1$. However, this is very rough reasoning, so don't trust me on that. — Noble Mushtak, Dec 28 '18 at 21:07

score 5 · Answer 1 · answered Dec 28 '18 at 21:41

5

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 $1+O(\frac{1}{\ln(n)})$.

answered Dec 28 '18 at 21:41

A. Pongrácz

7,418

Thank you fo helping!!! – M.Martinez Dec 29 '18 at 00:28
1

I am glad I could help. If you think this answers your question, you should accept the answer. – A. Pongrácz Dec 29 '18 at 07:45

Compute $\lim_{n\to\infty}\frac{\tfrac{n}{1}+\tfrac{n-1}{2}+\dots+\tfrac{2}{n-1}+\tfrac{1}{n}}{\ln(n!)}$

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