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.

positiveintegers, this is just a Dirichlet series in disguise. Successively take the limit, as $\Re z\to\infty$, of $T(z)$, $2^zT(z)$, $3^zT(z)$, etc.; the first limit that is nonzero will be $q_0^z T(z)$ (and the limit will equal $1$). $\endgroup$ – Greg Martin May 18 '18 at 21:00