# 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?

Take $$q_0 positive integers, $$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.

• Assuming you mean positive integers, 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 $1$). – Greg Martin May 18 '18 at 21:00
• @GregMartin: Fine. Thanks. I am thinking now if it can be merged into one formula without testing it all one by one. – alex.peter May 18 '18 at 21:03
• @GregMartin Something like create $g(x)=\lim\limits_{z \to \infty} x^zT(z)$ This would essentially extract all coefficients. So I guess the second part would be to detect the smallest coefficient, but if $x$ is binary position that means detecting first $1$ in binary notation. Hm, I will think about it. – alex.peter May 18 '18 at 21:13
• A formula would be $q_0 = \lim_z\rightarrow\infty T(z)^{-1/z}$. – Jan-Christoph Schlage-Puchta May 19 '18 at 13:16
• @Jan-ChristophSchlage-Puchta That much is derived already in the accepted answer. An improvement would be to get rid of the requirement $z>1$ for example, although it creates no difference in what I needed it for. – alex.peter May 19 '18 at 14:00

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}