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If $Re(s) > 1$ why $\log(\zeta(s)) = -\sum_{p \in \mathcal{P}} \log(1 - 1/p^s)$ ?

I know that $\zeta(s) = \prod_{p} (1 - 1/p^s)^{-1}$ and my book says that if $|z| < 1$ then $\log(1 - z) = \sum_{n \geq 1} z^n/n$

But if I take for instance : $a = 0.311+0.485i$ ; $b = 0.674+0.703i$ ; $c = 0.963+0.244i$

Then $\log((1-a)(1-b)(1-c)) \sim -1.826+3.113i$

and $\log(1-a) + \log(1-b) + \log(1-c) \sim -1.826-3.170i$

So how to show properly the property?

Gary
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badinmaths
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1 Answers1

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Just use the logarithmic product rule $\ln(ab)=\ln(a)+\ln(b)$ and $\ln(a^{-1})=-\ln(a)$ to get that (applied to finite product and then taking the limit and using continuity), that $\zeta(s)=\prod_{p\in P} (1-p^{-s})^{-1}$ implies for $\Re(s)>1$, that $\ln(\zeta(s))=\sum_{p\in P} \ln( (1-p^{-s})^{-1})= -\sum_{p\in P} \ln (1-p^{-s})$ which finishes the proof.

Eric
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