Take $q_0<q_1<...<q_k<q_{k+1}<...$ positive integers, $z$ complex
From
$$T(z)=\sum\limits_{k=0}^{\infty}\frac{1}{q_k^z}$$
I would need to extract the first coefficient $q_0$
It is irrelevant if this procedure is not achievable in practice. I would like to know if there is a principal solution. We do not know anything special about $q$'s. They could be just any positive integers.
I feel that the coefficient would have to appear somewhere, but where?
EDIT:
Now I can explain the reason for this. I was trying to make something out of Euler function and Riemann to make the used sieving more explicit.
$$E_1(s)=\sum\limits_{n=1}^{\infty}\frac{1}{n^s}=\zeta (s)$$
Let us create a function that is capable of extracting the smallest coefficient from the reciprocal sum of any ordered integers
$$f(s)=\sum\limits_{k=1}^{\infty}\frac{1}{q_k^s}$$
(this is where the answer fits)
$$ \lim\limits_{s \to \infty} \frac{1}{f(s)^\frac{1}{s}}=q_1$$
We have
$$E_{n+1}(s)=(1-\frac{1}{p_n^s})E_n(s)$$
which is eliminating all coefficients divisible by $p_n$.
$n^{th}$ prime number $p_n$ is then simply
$$ p_n=\lim\limits_{|s| \to +\infty} \frac{1}{(E_n(s)-1)^\frac{1}{s}}$$
I hope others will find this at least a little bit interesting as well.
$$ \ln(p_n)=-\lim\limits_{|s| \to +\infty} \frac{\ln\left (\zeta(s)\prod\limits_{k=1}^{n-1}(1-p_k^{-s})-1 \right )}{s} $$
Essentially the explicit form of the selection of the next prime is not given as we feel that is kind of obvious. Still, I find it missing in the sieving logic of the Euler product formula.