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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.

1 Answers 1

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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

  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)
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    @Chris Thus $L=\lim_{n\to\infty}(1+\gamma_n/E_n)^{E_n}=\lim_{n\to\infty}\left((1+\gamma_n/ E_n)^{E_n/\gamma_n}\right)^{\gamma_n}$ where $E_n=(\ln n!)/n$ and $\gamma_n\to\gamma$, Q.E.D.2012-06-14