Question
Let $p_r$ be the $r'th$ prime. Is it true that,
$$\sum_{r=1}^\infty s^r \ln(p_r) \sim \frac{s}{(1-s)} $$
I know this looks bizarre but kindly consider the argument below. I'm also interested in knowing if this technique (of summing to simplify via asymptotics) is known?
Argument
I recently observed an interesting behavior of the following series:
$$ K(s) = s \ln(2) + s^2 \ln(3) + s^3 \ln(5)+ \dots = \sum_{r=1}^\infty s^r \ln(p_r)$$
where $p_r$ is the $r'th$ prime
$$ \implies \lim_{n \to \infty} \sum_{r=0}^n s^r K(s) = \sum_{r=1}^\infty s^r \ln(\# p_r)$$
where $\# p_r = 2 \times 3 \times \dots \times p_r$
$$ \frac{K(s)}{1-s} \sim \sum_{r=1}^\infty s^r r = \frac{s}{(1-s)^2}$$
Where $s \nearrow 1$
$$ \implies K(s) \sim \frac{s}{(1-s)}$$