# How does one prove $\ln \zeta (s)=\sum _{{n=2}}^{\infty}{\frac{\Lambda (n)}{\log(n)}}\,{\frac{1}{n^{s}}}$

Introduction

The Von Mangoldt function is defined as follows: $$\Lambda (n)={\begin{cases}\log p&{\text{if }}n=p^{k}{\text{ for some prime }}p{\text{ and integer }}k\geq 1,\\0&{\text{otherwise.}}\end{cases}}$$

On the Wikipedia page over the Von Mangoldt function the identity below is listed: $$\ln\zeta(s)=\sum _{{n=2}}^{\infty}{\frac{\Lambda (n)}{\ln(n)}}{\frac{1}{n^{s}}}\qquad {\text{Re}}(s)>1$$ Question

How do you prove this? I'm not familiar with the Von Mangoldt function so pardon me if the identity has a completely trivial derivation, but I couldn't find it anywhere so I had to ask here.

If $$n=p^k$$ with $$p$$ prime and $$k\geqslant 1$$ then $$\frac{\Lambda(n)}{\log(n)}=\frac{\log(p)}{\log(p^k)}=\frac{1}{k}$$, thus $$\sum_{n=1}^{+\infty}\frac{\Lambda(n)}{\log(n)n^s}=\sum_{p^k}\frac{1}{kp^{ks}}$$ Moreover, the Euler's product formula for $$\zeta$$ states that $$\zeta(s)=\prod_{p\text{ prime}}\left(1-\frac{1}{p^s}\right)^{-1}$$ Thus $$\log\zeta(s)=-\sum_{p\text{ prime}}\ln\left(1-\frac{1}{p^s}\right)=\sum_{p\text{ prime}}\sum_{k=1}^{+\infty}\frac{1}{kp^{ks}}=\sum_{p^k}\frac{1}{kp^{ks}}=\sum_{n=1}^{+\infty}\frac{\Lambda(n)}{\log(n)n^s}$$
• Thanks for the answer. Just a quick question, how does $$-\sum_p\ln\left(1-\frac{1}{p^s}\right) = \sum_p\sum_{k=1}^{\infty}\frac{1}{kp^{ks}}$$ Are you using the taylor series for $\ln(x+1)$ or? Apr 28, 2020 at 14:33
• Yes, I'm using $$\ln(1-x)=-\sum_{k=1}^{+\infty}\frac{x^k}{k}$$ Apr 28, 2020 at 14:34