0

Consider the following zeta function well defined for $\operatorname{Re}(s) > 0$

$$\zeta_p(s) = 1 - \sum_{n=1}^{\infty} \frac{(-1)^n}{p_n^s}$$

where $p_n$ is the $n$ th prime.

So for clarity

$$\zeta_p(s) = 1 + \frac{1}{2^s} - \frac{1}{3^s}+ \frac{1}{5^s} - \frac{1}{7^s} +... $$

Is the following conjecture true ?

$$\zeta_p(s) = 0 \implies \operatorname{Re}(s) \leq \frac{3}{4}$$

edit : the below part is disproven now by Gary who linked the digits :

https://oeis.org/A078437

wrong

Also is it true that

$$ \zeta_p(1) = \frac{27}{32} + \frac{\pi}{8}+\frac{5}{48 \pi} $$

wrong

If one of them gets answered I might split up this question so that I can accept 2 answers.

edit : I split the question :

Is this prime identity true : $\frac{27}{32} + \frac{\pi}{8}+\frac{5}{48 \pi} = 1 - \sum_{n=1}^{\infty} \frac{(-1)^n}{p_n}$?

mick
  • 17,886
  • $\zeta_p(1)$ has been checked with over $1000000$ primes. – mick Jun 30 '25 at 00:37
  • Maybe useful : https://math.stackexchange.com/questions/4956800/bounds-for-the-prime-zeta-function – mick Jun 30 '25 at 00:58
  • 3
    By A078437, $$ \zeta _p (1) - \left( {\frac{{27}}{{32}} + \frac{\pi }{8} + \frac{5}{{48\pi }}} \right) = - 9.53 \ldots \times 10^{ - 9} \neq 0. $$ – Gary Jun 30 '25 at 01:01
  • Thanks @Gary that solves half the question ! – mick Jun 30 '25 at 01:15
  • @Gary https://math.stackexchange.com/questions/5079555/is-this-prime-identity-true-frac2732-frac-pi8-frac548-pi Plz insert the answer there, I will accept. – mick Jun 30 '25 at 01:29

1 Answers1

5

We can prove fairly easily that $\zeta_p(s)$ has infinitely many zeroes with $\Re s >1$.

Fix $a>0$ and since $\sum p^{-1-a}$ converges, we can choose $m$ st $1+\sum_{p \le m}p^{-1-a}>\sum_{p>m}p^{-1-a}$ and let $a(n)=1, n \le m$ and $a(n)=-1, n >m$ and consider $$Z_a(s)=1+\sum a(p)p^{-s}$$ Clearly as $s \to 1^+$ we have $Z_a(s) \to -\infty$ since $\sum p^{-1}=\infty$ while by choice we have $Z_a(1+a)>0$ so $Z_a$ has a zero in between $1$ and $1+a$

Now let's fix such an $a$ take $c>1$ a zero of $Z_a$ between $1$ and $1+a$ and pick $0 <\eta <c-1$ small enough st $Z_a(s)$ has no zeroes on $|s-c|=\eta$ and take $0<\epsilon=\min_{|s-c|=\eta}|Z_a(s)|$

Note that by choice $c-\eta>1$ so $1+\sum_p p^{-c+\eta}<\infty$ so we can find $P$ st $\sum_{p >P}p^{-c+\eta}<\epsilon/4$ and let $1+\sum_{p \le P}p^{-c+\eta}=A$

Now $\log p$ are linearly independent over the rationals so by Kronecker approximation theorem, we can find $t$ st $|p_n^{it}-(-1)^{n+1}a(p_n)|<\epsilon/(2A)$ for all $p_n \le P$ so on $|s-c|=\eta$ we have by the triangle inequality and our choices $$|\zeta_p(s+it)-Z_a(s)|\le \frac{\epsilon}{2A}\sum_{p \le P}p^{-\Re s}+2\sum_{p>P}p^{-\Re s}$$

But $\Re s \ge c-\eta$ on $|s-c|=\eta$ so $p^{-\Re s} \le p^{-c+\eta}$ there so $$|\zeta_p(s+it)-Z_a(s)|< \frac{\epsilon}{2A}(A-1)+2\epsilon/4 < \epsilon $$ hence by Rouche $\zeta_p(s+it)$ has at least a zero in $|s-c|=\eta$ since $Z_a$ does. In particular for every zero $1<c<1+a$ and every $\eta>0$ small enough we have a zero $s_{c, \eta}$ of $\zeta_p$ with $\Re s_{c, \eta} \in [c-\eta, c+\eta]$. Since we can choose $c \to 1$ by taking $a \to 0$, we can also choose $\Re s_{c, \eta} \to 1$ so we have infinitely many such zeroes of $\zeta_p$ with $\Re s >1$

Conrad
  • 31,769
  • Arent you forgetting the part with $(-1)^n$ ? You seem to work with a pole at $1$ but my function has no such pole. And all seems based on that. I see your post as “ there could be “ rather than “there must be “. But maybe I’m just not getting it. That being said I wonder about examples of the zero’s. For $Re(s) > 1$ and $3/4 < Re(s) < 1 $. – mick Jun 30 '25 at 16:55
  • 2
    if you read carefully you will see that is included in the Dirichlet approximation theorem where we get $(-1)^na(p_n)$ approximated as closely as we want for as many primes as we want by $p_n^{it}$ for a single $t$; the existence or not of a pole is irrelevant since $Z_a$ approximates a translate of $\zeta_p$ to the right of $1+\eta$ where both functions are analytic - by same Kronecker approximation theorem it is easy to show that while $\zeta_p$ is analytic up to $0$ it is quite unbounded (vertically) near $\Re s=1$ so a vertical translate ican (and will) get close to $Z_a$ – Conrad Jun 30 '25 at 17:12
  • 2
    This is a standard proof with which you can show for example that any $\sum c_p p^{-s}$ with say $|c_p|=1$ (or more generally with $\sum |c_p/p|=\infty$) has infinitely many zeroes to the right of $1$ regardless if it converges to $\Re s=0$ say like in this case - more generally it applies to $\sum c_n g_n^{-s}$ where $g_n$ positive and increasing to infinity st $\log g_n$ are linearly independent over the rationals and $\sum |c_n/g_n|=\infty$ again regardless of where the series converges - eg the alternating Hurwitz zeta with transcendental offset $\sum \pm (n+a)^{-s}, a$ transcendental – Conrad Jun 30 '25 at 17:21
  • do you have free references to those things ? I am unfortunately unfamiliar with it and I should be familiar. Does your result say anything about $3/4 < Re(s) < 1 $ ?? I found some dirichlet approximation theory but it was not about exp sums or maybe i missed something. – mick Jun 30 '25 at 20:31
  • 2
    the standard reference is Davenport and Heillbronn 1936 paper On the Zeroes of Certain Dirichlet Series - see https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms/s1-11.3.181 - in which the above method is introduced; as for extending this type of density results to $\Re s <1$ (real part of zeroes dense in $[0,1]$ too) I see no reason for the above not to work with more care – Conrad Jun 30 '25 at 20:48
  • does this (in your opinion perhaps this is not formally answerable) imply that the only interesting zeta functions are those with an euler product ? If not what are some typical other interesting ones, and where are their zero's ( conjectured or not) ? – mick Jun 30 '25 at 20:57
  • but that is not free :/ – mick Jun 30 '25 at 21:19