Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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. – qgp07 Mar 31 '14 at 14:35
up vote 6 down vote accepted

It telescopes after you do a little algebra:

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

share|cite|improve this answer
Very clever! Thanks so much for your help. – James Evans Dec 10 '12 at 5:06
@dirk5959: You’re very welcome. – Brian M. Scott Dec 10 '12 at 5:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.