Good Morning everybody,

I'd like to show the equality

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

By logarithmic differenciation(using Euler product for zeta), I get

$\displaystyle \frac{\zeta'(s)}{\zeta(s)}=-\sum_p (\ln(1-p^{-s}))'=-\sum_p \frac{p^{-s}\ln p}{1-p^{-s}}$

But I can't figure how to reach the last step, to the double sum.

Could someone help me ?

Thank you