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

This is surely a tiny question but I seem to have some blackout currently ...
I tried to define a function for the sum of logarithms, like we have it for the sums of like powers with the bernoulli-polynomials. (I had a question with sums of logarithms here on MSE earlier, but it is not directly translatable). I've got for the following sum of logarithms

$ \qquad \small \sum_{k=a+1}^b \log(1+1/k)) $

the equivalent expression:

$ \qquad \small (\log(1/a)-\log(1+1/a) - ( \log(1/b)-\log(1+1/b)) $

but don't see, why.... This must have to do something with telescoping, but I just don't get it...

(The functions for the sums of the higher powers of the logarithms require series involving zetas as expected, so this simple contraction of a formula was extremely surprising)

share|cite|improve this question
up vote 9 down vote accepted

$\log(1 + 1/k) = \log(\frac{k + 1}{k}) = \log(k + 1) - \log(k)$

share|cite|improve this answer
Oh I forgot... "HINT"! – The Chaz 2.0 Sep 22 '11 at 22:15
arrggh... sometimes things are so simple... So my function was configured correctly; thanks! – Gottfried Helms Sep 22 '11 at 22:15
Glad to help :) – The Chaz 2.0 Sep 22 '11 at 22:16
This is used in deriving an infinite series for the Euler-Mascheroni constant. – anon Sep 22 '11 at 22:57
@anon: this page has become impressive, meanwhile. Thanks for the reminder... – Gottfried Helms Sep 23 '11 at 17:33

Write $$ \sum_{k=a+1}^b\log(1+1/k)=\log\left(\prod_{k=a+1}^b\frac{k+1}{k}\right) $$ or $$ \sum_{k=a+1}^b\log(1+1/k)=\sum_{k=a+1}^b\log(k+1)-\log(k) $$ Then you can work with a telescoping product or sum.

share|cite|improve this answer

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.