1
$\begingroup$

Consider the following partial sum:

$$S(x,n)=\sum_{p\leq x}\frac{\ln(p)}{({p})^{n/2}}$$

Here p runs through primes and $n$ is constant

What is the best possible unconditional( using best known version of PNT) estimate of the given sums?

I think for $n>2$ the sum converges for $x=\infty$.

Also, I think $n=1$ is the most crucial case

Improve this question
$\endgroup$
1
  • $\begingroup$ Your question seems related to the first Chebyshev function $\vartheta(x)=\sum\limits_{p\le x} \log(p)$ (see en.wikipedia.org/wiki/Chebyshev_function) and $\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{p\le N} \frac{log(p)}{p^s}\right)$ does indeed converge for $\Re(s)>1$. $\endgroup$
    – Steven Clark
    Commented Feb 6, 2023 at 15:12

1 Answer 1

Reset to default
3
$\begingroup$

Write

$$\theta(x) = \sum_{p\leq x}\log p.$$

By partial summation, you can show that

$$S(x,n) = \frac{\theta(x)}{x^{n/2}}+\frac{n}{2}\int_{2}^{x}\frac{\theta(t)}{t^{1+\frac{n}{2}}}dt.$$

The strongest forms of the prime number theorem to date tell us that there exists a constant $c>0$ such that if $t\geq 2$, then

$$|\theta(t)-t|\ll \begin{cases} \sqrt{t}(\log t)^2&\mbox{if the Riemann Hypothesis is true,}\\ t\exp(-c (\log t)^{3/5}(\log\log(3t))^{-1/5})&\mbox{otherwise.} \end{cases}$$

(I think that $c=1/100$ is permissible.) You can now estimate $S(x,n)$ for any $n$ that you want. If $n\leq 2$ (in which case $\lim_{x\to\infty}S(x,n)=\infty$) and you care about the quality of the error terms, then it might matter whether or not you assume the Riemann Hypothesis.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .