Limit as $n\to\infty$ of $\frac{\frac{n}{1}+\frac{n-1}{2}+\frac{n-3}{3}+...+\frac{2}{n-1}+\frac{1}{n}}{\ln(n!)}$ The task is to get the limit below:
$$\lim_{n\rightarrow \infty}\frac{\frac{n}{1}+\frac{n-1}{2}+\frac{n-3}{3}+\cdots+\frac{2}{n-1}+\frac{1}{n}}{\ln(n!)}$$
I used Stolz but I don't know how to subtract the sequence. 
 A: You can write the numerator as 
$$n(1 + 1/2 + \cdots +1/n) -(1/2 + 2/3 + 3/4 + \cdots (n-1)/n)$$
Note the first expression in parentheses is $\sim \log n.$ The second expression in parentheses is $O(n).$
The denominator is
$$\sum_{k=1}^{n} \ln k \sim \int_1^n \log x\, dx \sim n\log n.$$
The quotient thus looks like $(n\log n +O(n))/(n\log n) \to 1.$
A: By the Stolz-Cesaro theorem from http://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem, 
\begin{eqnarray}
&&\lim_{n\rightarrow \infty}\frac{\frac{n}{1}+\frac{n-1}{2}+\frac{n-3}{3}+\cdots+\frac{2}{n-1}+\frac{1}{n}}{\ln(n!)}\\
&=&\lim_{n\rightarrow \infty}\frac{\sum_{k=1}^n\frac{n+1-k}{k}}{\sum_{k=1}^n\ln k}\\
&=&\lim_{n\rightarrow \infty}\frac{\sum_{k=1}^{n+1}\frac{n+2-k}{k}-\sum_{k=1}^n\frac{n+1-k}{k}}{\sum_{k=1}^{n+1}\ln k-\sum_{k=1}^n\ln k}\\
&=&\lim_{n\rightarrow \infty}\frac{\sum_{k=1}^{n+1}\frac{1}{k}}{\ln(n+1)}\\
&=&\lim_{n\rightarrow \infty}\frac{\ln(n+1)+\gamma+o(1)}{\ln(n+1)}\\
&=&1.
\end{eqnarray}
A: An approach:
$$
\sum_{k=1}^n \frac{n-k+1}{k} = (n+1)\sum_{k=1}^n \frac{1}{k} - n = (n+1)H_n-n
$$
Now, that's the numerator;, and you also have $\log n! = \sum_{n=1}^n \log k = n\log n +o(n\log n )$ (not hard to show), while $H_n= \log n + O(1)$. Putting them together will give you the limit.
--
Edit: there was previously double counting for the numerator, removed one of the two sums (unnecessary).
A: You have 
$$\frac{\sum _{k=0}^{n-1} \left(\frac{k+1}{n-k}\right)}{\sum _{k=1}^{n}\log(k)}$$
After integral approximation:
$$\frac{\int_{-0.5}^{n-0.5} \left(\frac{k+1}{n-k}\right) \, dk}{\int^{n+0.5}_{0.5}\log(k)dk}$$
In the end the dominant parts of the ratios are 
$$\rightarrow\frac{n\log(n)}{n\log(n)}=1$$
A: Recall that $\ln(xy) = \ln(x)+\ln(y)$. So the bottom sequence actually looks like 
$$
\ln(1)+\ln(2)+...+\ln(n)
$$
So if $a_n = \sum_{i=1}^{n}\frac{i}{n-(i-1)}$, then 
$$
a_{n+1} -a_{n}= \sum_{i=1}^{n+1}\frac{i}{n+1-(i-1)}-\sum_{i=1}^{n}\frac{i-1}{n-(i-1)}= \sum_{i=1}^{n+1}\frac{1}{i}
$$
and $b_n = \ln(n!)$ so we have 
$$
b_{n+1}-b_n = \ln(n+1)
$$
The rest should be fairly standard.
Incidently, if you're interested in harmonic vs. logarithmic growth, check out the Euler–Mascheroni constant.
