1. ## [SOLVED] Logarithmic derivation

Good Morning everybody,
I'd like to show the equality
$\frac{\zeta'(s)}{\zeta(s)}=-\sum_p\sum_{m=1} \frac{\ln p}{p^{ms}}$
By logarithmic differenciation(using Euler product for zeta), I get
$\frac{\zeta'(s)}{\zeta(s)}=-\sum_p (\ln(1-p^{-s}))'=-\sum_p \frac{p^{-s}\ln p}{1-p^{-s}}$
But I can't figure how to reach the last step, to the double sum.
Could someone help me ?
Thank you

2. Originally Posted by Klaus
Good Morning everybody,
I'd like to show the equality
$\frac{\zeta'(s)}{\zeta(s)}=-\sum_p\sum_{m=1} \frac{\ln p}{p^ms}$
By logarithmic differenciation(using Euler product for zeta), I get
$\frac{\zeta'(s)}{\zeta(s)}=-\sum_p (\ln(1-p^{-s}))'=-\sum_p \frac{p^{-s}\ln p}{1-p^{-s}}$
But I can't figure how to reach the last step, to the double sum.
Could someone help me ?
Thank you

Shouldnt it read $\frac{\zeta'(s)}{\zeta(s)}=-\sum_p\sum_{m=1} \frac{\ln p}{p^{ms}}$ ??

You are nearly there

Can you see the geometric progression below?
$-\sum_p \frac{p^{-s}}{1-p^{-s}}=-\sum_p \sum_m (p^{-s})^m$

3. Oh yeah, that's it !
I hadn't thought of geometric progessions
Thank you very much Isomorphism(what a cool name!)