# The Half-Plane of Convergence of Prime Zeta Function

This post contains a brief proof of Conj. 3 posted on Lee’s Conjecture. To avoid confusion I used $x$ as real variable, and $s = \sigma + it$ as complex.

I proposed a conjecture as follows:

Conj. 3:
Let $\mathbb{P}$ be a set of primes. The supremum of $x$ such that $\displaystyle \sum_{p \in \mathbb{P}} \frac{1}{p^x}$ diverges is $x = 1$.

Let us define a Prime Zeta Function as follows: $\displaystyle P(s) = \sum_{p \in \mathbb{P}} \frac{1}{p^s}$.
In fact, the conjecture turned out to be an easy comparison series problem.

Consider the original Riemann Zeta function, that is defined as $\displaystyle \zeta(s) = \sum_{n \in \mathbb{N}} \frac{1}{n^s}$. It was already proven that $\zeta(s)$ is defined on a half plane of $s$ such that $\sigma > 1$. The domain was extended over all complex plane except $s = 1$ via Analytic Continuation in Riemann’s paper.

Since the conjecture is asking the supremum of $x$, let us only consider the case of real numbers. Obviously, $\zeta(x) \geq P(x)$ for $x > 1$, as each term in the summand of $P(x)$ is also included in $\zeta(x)$, and every term in $\zeta(x)$ is positive. Thus, as long as $\zeta(x)$ is defined, so is $P(x)$. Hence, the domain of $P(x)$ includes a half real line $x > 1$. Therefore, the supremum of $x$ where $P(x)$ diverges must be $x = 1$; the conjecture is true.

While studying Introduction to Analytic Number Theory by T. Apostol, I encountered a following theorem. (1976 Edition, pp.233)

Theorem 11.8:
If [a Dirichlet] series $\sum f(n)n^{-s}$ converges for $s = \sigma_0 + it_0$, then it also converges for all $s$ with $\sigma > \sigma_0$. If it diverges for $s = \sigma_0+it_0$, then it also diverges for all $s$ with $\sigma < \sigma_0$

$f(n)$ presented in the theorem is an arithmetical function. Consider a prime detecting function $\rho (n)$ such that $\rho(n) = 1$ for $n$ being prime, and $\rho(n) = 0$ for $n$ being composite. Then, the Dirichlet series $\sum \rho(n)n^{-s} = P(s).$ Therefore, we may apply Theorem 11.8 to deduce the following fact.

We already showed that for any real number $x > 1, P(x)$ converges. Take arbitrarily small $\epsilon > 0$, then $P(1+\epsilon)$ converges. Thus, by Theorem 11.8, for any $s$ with $\sigma > 1+\epsilon, P(s)$ also converges. Letting $\epsilon \to 0$, we can conclude $P(s)$ is defined for all $\sigma > 1$. Thus, the domain of $P(s)$ is now extended to the half plane $s$ where $\sigma > 1$.

Then, we can come up with a few new problems.

1. Can we extend the domain of $P(s)$ via Analytic Continuation?
2. As we defined Prime Zeta Function, let us define Composite Zeta Function, that is $\displaystyle C(s) = \sum_{n \in \mathbb{N} \setminus \mathbb{P}} \frac{1}{n^s}$. Will the Composite Zeta Function have the same domain?
3. Does Prime Zeta Function have functional equation as Riemann Zeta Function? If so, what are the trivial zeroes and nontrivial zeroes of $P(s)$?