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In working through a problem, I've encountered the need to express

$\log n = \sum \limits_{k = 1}^{n - 1} \log(1 + \frac{1}{k})$

where $\log k $ is the natural logarithm of k. I'm fairly certain the above equality holds (please let me know if otherwise), but I'm not quite sure how to derive it. Any help would be greatly appreciated.

So far, I've tried expressing $\log n = \log(n!) - \log((n-1)!)$ to no avail.

Many thanks!

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  • $\begingroup$ You were really close to the answer you accepted; if you expanded $\log(n!)$ and $\log((n-1)!)$ separately and resummed them, you'd basically get the given answer in reverse. $\endgroup$ – qgp07 Mar 31 '14 at 14:35
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It telescopes after you do a little algebra:

$$\begin{align*} \sum_{k=1}^{n-1}\log\left(1+\frac1k\right)&=\sum_{k=1}^{n-1}\log\frac{k+1}k\\\\ &=\sum_{k=1}^{n-1}\big(\log(k+1)-\log k\big)\\ &=\log n-\log 1\\ &=\log n\;. \end{align*}$$

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  • $\begingroup$ Very clever! Thanks so much for your help. $\endgroup$ – James Evans Dec 10 '12 at 5:06
  • $\begingroup$ @dirk5959: You’re very welcome. $\endgroup$ – Brian M. Scott Dec 10 '12 at 5:07

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