asymptotic expansion in powers of $ 1/s $

Question

How could I get the coefficients in the Dirichlet series expansion

$$ \frac{\zeta ' (s)}{\zeta(s)}= \sum_{n=1}^{\infty} \frac{a(n)}{n^s} $$

for the logarithmic derivative of the Riemann zeta function?

Answer 1

Start with

$$\zeta(s) = \sum_{n=1}^{\infty} \frac1{n^s} $$

$$\zeta'(s) = -\sum_{n=1}^{\infty} \frac{\log{n}}{n^s} $$

Further,

$$\frac1{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} $$

where $\mu(n)$ is the Möbius function.

Then

$$\frac{\zeta'(s)}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{a(n)}{n^s} $$

where

$$a(n) = -\sum_{p q=n} \mu(p) \log{q} $$

and $p \in \mathbb{N}$, $q \in \mathbb{N}$