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.



Is there a procedure for extracting first integer $q_0$ from $\sum\limits_{k=0}^{\infty}\frac{1}{q_k^z}$, all $0<q_0<q_1<...$ integers, $z$ complex?
For $z>1$, we have 
\begin{equation}
 q\_0^{-z}\le T(z)\le\sum\limits\_{q=q\_0}^{\infty}q^{-z}\le q\_0^{-z}+\int\_{q\_0}^\infty q^{-z}\,dq
 =q\_0^{-z}(1+q\_0/(z-1)).
\end{equation}
Hence, 
\begin{equation}
 q\_0=\lim\_{z\to+\infty}T(z)^{-1/z}. 
\end{equation}


