real analysis - Computing: $L =lim_{nrightarrowinfty}left(frac{frac{n}{1}+frac{n-1}{2}+cdots+frac{1}{n}}{ln(n!)} right)^{{frac{ln(n!)}{n}}} $

Compute the following limit:

$$L =\lim_{n\rightarrow\infty}\left(\frac{\frac{n}{1}+\frac{n-1}{2}+\cdots+\frac{1}{n}}{\ln(n!)} \right)^{{\frac{\ln(n!)}{n}}} $$

I'm looking for an easy, simple solution here, but not sure yet this is possible. Any hint, suggestion along this way is welcome. Thanks.

Answer

I'm not sure that I'm right.

First we have
$\sum_{k=1}^n (n+1-k)/k = (n+1)H_n-n$,
so
$$L = \lim_{n\to\infty} \left(\frac{(n+1)H_n-n}{\ln n!}\right)^{\frac{\ln n!}n}$$
Take logarithm, we have $\ln L = \lim_{n\to\infty} A(n)B(n)$, where
$$A(n) = \frac{\ln n!}n = \frac{n\ln n+O(n)}n = \ln n+O(1)$$
and $B(n) = \ln C(n)$ where
\begin{align*}
C(n)
&= \frac{(n+1)H_n-n}{\ln n!} \\

&= \frac{(n+1)(\ln n+\gamma+O(1/n))-n}{n\ln n-n+O(\log n)} \\
&= \frac{n\ln n-(1-\gamma)n+O(\log n)}{n\ln n-n+O(\log n)} \\
&= \frac{1-\dfrac{1-\gamma}{\ln n}+O(1/n)}{1-\dfrac1{\ln n}+O(1/n)} \\
&= \left(1-\frac{1-\gamma}{\ln n}\right)\left(1-\frac1{\ln n}\right)^{-1}\left(1+O(1/n)\right)^2 \\
&= \left(1-\frac{1-\gamma}{\ln n}\right)\left(1+\frac1{\ln n}+O(1/\log n)^2\right)\left(1+O(1/n)\right) \\
&= 1+\frac\gamma{\ln n}+O(1/\log n)^2
\end{align*}
So
$$B(n) = \ln C(n) = \ln\left(1+\frac\gamma{\ln n}\right)+O(1/\log n)^2 = \frac\gamma{\ln n}+O(1/\log n)^2$$
and

$$A(n)B(n)=\gamma+O(1/\log n)$$
Let $n\to\infty$, we have $\lim_{n\to\infty} A(n)B(n)=\gamma$, so $L = e^\gamma$.

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The following equations come from Concrete Mathematics, proved by Euler-Maclaurin formula

1. $H_n = \sum_{k=1}^n 1/k = \ln n+\gamma+O(1/n)$, where $\gamma$ is Euler-Mascheroni constant.


2. $\ln n! = n\ln n-n+O(\log n)$. (It's really Stirling's approximation)