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.


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)
  • $\begingroup$ This looks allright, except you forgot to transform $1-\frac1{\log n}$ in the denominator into $1+\frac1{\log n}$ in the numerator when computing $C(n)$. Hence $L=e^\gamma$. $\endgroup$ – Did Jun 13 '12 at 14:34
  • $\begingroup$ @did Thanks. I've been too nervous. I'll try to edit it. $\endgroup$ – Yai0Phah Jun 13 '12 at 14:38
  • $\begingroup$ Nice solution! $$ $\endgroup$ – user17762 Jun 14 '12 at 8:28
  • $\begingroup$ @Chris Since the limit depends on $\gamma$, I guess it is hard to find other easier ways to do this. To evaluate this limit, at some point, you need to know an asymptotic for the harmonic number. $\endgroup$ – user17762 Jun 14 '12 at 8:31
  • $\begingroup$ @Chris $L=\lim_{n\to\infty}\left(1+\dfrac{(H_n-1-(\ln n!)/n)+H_n/n}{(\ln n!)/n}\right)^{(\ln n!)/n}$. $\endgroup$ – Yai0Phah Jun 14 '12 at 9:21

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