Understanding a sum with logs and prime numbers.

Question

I'm reading a proof and I don't understand this step:

$q$ are the primes

$$-\sum_q \ln \left(1-\frac{1}{q^s}\right) = \sum_q \frac{1}{q^s} + \mathcal{O}(1)$$

Working with

$$-\ln \left(1-\frac{1}{q^s}\right) = -\ln \left(\frac{q^s-1}{q^s}\right)=\ln \left(\frac{q^s}{q^s-1}\right)=s\ln (q)+ \ln\left(\frac{1}{q^s-1}\right)$$

I've been messing with it for a bit and not seeing the connection.

Answer 1

We have $$\begin{align}-\sum_q \ln\left(1-\frac{1}{q^s}\right)&=\sum_q\sum_{k=1}^\infty\frac1{kq^{ks}}\\&\le\sum_q\left(-1+\sum_{k=0}^\infty\left(\frac1{q^s}\right)^k\right)\\&=\sum_q\left(-1+\frac1{1-\frac1{q^s}}\right)\\&=\sum_q\frac1{q^s-1}\\&=\sum_q \frac{1}{q^s} + \mathcal{O}(1)\end{align}$$