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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.

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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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  • $\begingroup$ 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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