Simplifying $\frac{\zeta'(s)}{\zeta(s)}$

Question

$$\text{I need to prove the following lemma : }\frac{\zeta'(s)}{\zeta(s)} = - \sum_{n=1}^{\infty} \frac{\Lambda(n)}{n^s}$$

My attempt:

$$\text{We know that }\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s} $$$$\text{So, }\zeta'(s) = \sum_{n=1}^{\infty}\frac{d}{ds}.\frac{1}{n^s} = -\sum_{n=1}^{\infty} \frac{ln(n)}{n^s}$$

This gives : $$\frac{\zeta'(s)}{\zeta(s)} = -\frac{\sum_{n=1}^{\infty}\frac{1}{n^s}}{\sum_{n=1}^{\infty} \frac{ln(n)}{n^s}} $$

How do I proceed ? Please help.

Answer 1

It is easy to show that $\log n = \sum_{d | n } \Lambda(d)$. So by Mobius inversion it follows that $\Lambda(n) = \sum_{d|n} \mu(d) \log(n/d)$. We know that $$\frac{1}{\zeta(s)} = \sum _{n=1}^{\infty} \frac{\mu(n)}{n^s}$$ and $$\zeta'(s) = -\sum_{n=1}^{\infty} \frac{\log n}{n^s}$$, so by Dirchlet convolution we have $$\frac{\zeta'(s)}{\zeta(s)} = - \sum_{n=1}^{\infty} \sum_{d|n} \mu(d) \log(n/d) \frac1{n^s} = -\sum_{n =1}^{\infty} \frac{\Lambda(n)}{n^s}$$

Answer 2

HINT:$$\frac{\zeta'(s)}{\zeta(s)}=\frac{d}{ds}\ln\zeta(s)=-\sum_p\frac d{ds}\ln(1-p^{-s}).$$