Computing: $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}}} $ 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.
 A: 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 $L = e^\gamma$.

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


*

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

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

