# how to prove this extended prime number theorem?

A Generalized Prime Number Theorem?

Conjecture

Let $n$ and $k$ be positive integers with $n - 50 > k^2 > 0$ and $n$ sufficiently large. Then for the odd primes we have, when $p$ is the biggest odd prime $\le n$, $$3^k + 5^k + 7^k + 11^k + ... + p^k \sim \frac{n^{k+1}}{(k+1) ( \log(n) - \log(k) ) }$$ I wonder if you guys have seen it before ?

How to prove it ?

Any useful references for $k > 1$ ?

• Remark: Also posted on Math Overflow, but the question fits better here. – Eric Naslund Sep 3 '12 at 22:58
• Possible duplicate of How does $\sum_{p\leq x}p^{-s}$ grow asymptotically for $\text{Re}(s)<1$? The answer there proves the above uniformely for $k\leq e^{c\sqrt \log x}$. – Eric Naslund Sep 3 '12 at 23:02
• You write, "Theorem" and "we have...," which is what mathematicians write about something that has been proved. Then you ask how to prove it. Please edit your question to make it clear: is this something you have proved, or is it something you can't prove? – Gerry Myerson Sep 3 '12 at 23:11
• I slightly edited it. – mick Sep 4 '12 at 8:55

For $k<e^{c\sqrt{\log x}}$, you can prove the above using partial summation along with the prime number theorem, but it is provably false for $k\sim x^{u}$ when we assume RH. For a complete solution, take a look at this past answer. Specifically, I prove that $$\sum_{p\leq x}p^{k}=\text{li}\left(x^{k+1}\right)+O\left(x^{k+1}E(x)\right),$$ where $\text{li}(x)=\int_2^x \frac{1}{\log t}dt$ is the logarithmic integral, and where $E(x)$ is any positive increasing function which bounds the error term $\pi(x)-\text{li}(x)$. This gives the asymptotic $$\sum_{p\leq x}p^{k}\sim \frac{x^{k+1}}{(k+1)\log x}$$ uniformely for all $k<e^{c\sqrt{\log x}}$, and if you use the Walfisz bound it can be taken slightly further to $e^{c(\log x)^{3/5}}$ with some doubly logarithmic terms in the exponent.

If you assume the Riemann Hypothesis, this shows that your above expressions is not correct for $k\approx x^u$ where $0<u<1$. On RH we have that $E(x)\ll x^{\frac{1}{2}+\epsilon}$, and so the asymptotic holds for all $k<x^{\frac{1}{2}-\epsilon}$. Then for $k=x^u$ we have $\log x-\log k=(1-u)\log x$, which yields the asymptotic $$(1-u)\frac{x^{k+1}}{(k+1)},$$ which is off by the constant factor $(1-u)$.

There may be a simpler way to prove that your expression is incorrect for $k\sim x^u$ that does not require RH. I think you need to be clever though.

Hope that helps,

• Hmm i wonder why i got the factor (1-u) then. Is it possible that the factor (1-u) fits better for small x and fails for large x ? Could i have made roundoff errors or to brute estimates ? I need some time to think about this. Im a bit confused :)Thanks. – mick Sep 4 '12 at 8:50

A slightly weaker form is the direct application of:

Theorem: Let $$p_k$$ be the $$k$$-th prime and let $$f$$ be a continuous function Riemann integrable in $$(0,1)$$ then,

$$\lim_{m \to \infty}\frac{1}{m}\sum_{r = 1}^{m}f\Big(\frac{p_r}{p_m}\Big) = \int_{0}^{1}f(x)dx.$$

For $$f(x) = x^k$$ we get the more elegant form

$$2^k + 3^k + 5^k + \ldots + p_m^k \sim \frac{mp_m^k}{k+1} \sim \frac{m^{k+1} \log^k m}{m+1}$$

and then express it in terms $$n$$ using $$m \log m \sim n$$.