0

I am curious about the below function, as it allows one to build a perfect ordering of the prime numbers through a method that I developed, that is, there is a function $f(n)$ such that $f(n)$ is the n-th prime.

Let’s call this new zeta function $\zeta^*$:

$$\zeta^*(s)=\sum_{k=1}^{\infty}\frac{p_k}{k^s}$$

Where $p_k$ is obviously the k-th prime number.

Now the questions that naturally arise are:

1) what is the smallest positive $s>1$ for which this function converges?

2) what are the values of this function at the positive integers?

Any other information about this function helps as well.

2 Answers2

1

The explicit formula for $\pi(x)=li(x)-\sum_\rho li(x^\rho)$ translates to an explicit formula for $p_k$ which won't be a sum but a more complicated thing with non-linear terms. Nevertheless you can connect both and find that $\sum_k \frac{p_k}{k^s}$ encodes the primes and the zeta zeros in a very similar way to $\sum_k \frac{\pi(k)}{k^s}$ thus to $\log \zeta(s-1)$.

$\sum_k \frac{p_k}{k^s}$ has complicated branch points at $s=2$ and $s=\rho+1$ and an analytic continuation to $\Re(s) > 3/2$

It converges for $\Re(s)> 2$, it diverges for $\Re(s) < 2$ and the PNT in its form $\pi(x)=li(x)+O(x/\log^3 x)$ gives $p_k=n\log n+a n+bn/\log n+O(n/\log^2 n)$ indicating that it diverges on $\Re(s)=2$.

It is hopeless to expect closed-form special values at integers: the function $k\to p_k$ doesn't have any practical generating function as $\log \zeta(s)$.

reuns
  • 77,999
0

By the prime number theorem, $\pi(x) \sim x / \ln x$ and $p_k \sim k \ln k$, so by Cauchy's root test:

$\begin{align*} \lim_{k \to \infty} \sqrt[k]{p_k / k^s} &= \lim_{k \to \infty} \sqrt[k]{k \ln k / k^s} \\ &= \lim_{k \to \infty} \sqrt[k]{k^{1 - s} \ln k} \ \end{align*}$

To simplify the root, take logarithms:

$\begin{align*} \lim_{k \to \infty} \frac{(1 - s) \ln k + \ln \ln k}{k} \end{align*}$

This limit is zero for all $s$, thus this test is inconclusive (the original limit is 1). Thus the ratio test is also inconclusive.

Would need to check other tests, but I ran out of steam.

vonbrand
  • 27,812