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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)

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2 Answers

up vote 9 down vote accepted

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

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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
2  
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
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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.

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