0

Is it known wether the following limit tends to Infinity or not? Is there any possibility for it to converge to a constant?

$$\lim_{n \to \infty} \sum^n_{p\le n} \frac{1}{\sqrt{p}} - li(\sqrt{n})$$

(Being $p$ a prime number and $li(k)$ the logarithmic integral)

I have not found anything about it. It seems that most efforts are made trying to upper-bound the asymptotic behaviour of the limit. For example, here there is a proof of this behaviour based on the Prime Number Theorem. Also, the Riemann's Hypothesis provides a tighter upper bound. But what about a lower bound?

Community
  • 1
user3141592
  • 1,859
  • Duplicate http://math.stackexchange.com/questions/2198027/limit-involving-sum-over-prime-numbers if you don't understand something say it. As I said this is a complicated subject. Read https://en.wikipedia.org/wiki/Dirichlet_series#Abscissa_of_convergence and use the properties of $\int_2^\infty li(x^{1/2}) x^{-s-1}dx$, you'll see that if $\sum^n_{p\le n} \frac{1}{\sqrt{p}} - li(\sqrt{n})$ converges then $\sum_p p^{-s}- \log(s-1)$ is analytic on $Re(s) > 1/2$ so that $\log \zeta(s) = \sum_{p^k} \frac{p^{-sk}}{k}$ is analytic and $\zeta(s)$ has no zeros there (the Riemann hypothesis) – reuns Mar 26 '17 at 16:58
  • I know that, it the limit converges, it would imply the RH. But if the RH were proven to be true, this limit wouldn't necessarily converge, isn't it? Hence my question – user3141592 Mar 26 '17 at 17:03
  • Start reading a proof of the prime number theorem. That RH $\implies \lim_{n \to \infty}\sum^n_{p\le n} \frac{1}{\sqrt{p}} - li(\sqrt{n})$ converges is very a complicated question. – reuns Mar 26 '17 at 17:05
  • 1
    Also can you relate $li(\sqrt{x})$ with $\sum_{n = 2}^x \frac{1}{\sqrt{n}\ln n}$ ? And you didn't answer : what do you want to do ? Plotting something related to the Riemann hypothesis ? – reuns Mar 26 '17 at 17:07
  • @user1952009 I decided to ask this as, if you follow the answer of the link I attached, and instead of using the error term of that PNT you use the asymptotic $O(n^{1/2} \log n)$, and then set $s=1/2$, I am not able to get to the convergence of this limit – user3141592 Mar 26 '17 at 17:13

1 Answers1

1

In your linked question it is using the prime number theorem, that is equivalent to $$\sum_{p < x} p^{-1/2} - li(\sqrt{x}) = o\left(\frac{x^{1/2}}{(\ln x)^2}\right)$$ That's all we can do, the PNT has been proven in 1900 and no major step has been done until today.

It is already quite hard to show that the Riemann hypothesis is equivalent to $$\forall \epsilon > 0, \qquad\sum_{p < x} p^{-1/2} - li(\sqrt{x}) = o\left(x^{\epsilon}\right)$$

Assuming the RH the experts can do better, obtaining $O(\ln^2 x)$ as I mentioned in your other question. If you want to know the best error term, make some research you'll find easily many papers about that.

Community
  • 1
reuns
  • 77,999
  • or $o\left(\frac{x^{1/2}}{(\ln x)^k}\right)$ if you want a better error term – reuns Mar 26 '17 at 17:15
  • Ok, I got that. My problem is when trying to use the RH instead – user3141592 Mar 26 '17 at 17:16
  • 1
    @user3141592 This is a too difficult subject for you, and I won't search 2 hours for helping you if you don't explain what you want really – reuns Mar 26 '17 at 17:16
  • Ok, I understand. Sorry – user3141592 Mar 26 '17 at 17:20
  • 1
    To be fair, this is a very difficult problem! However, it is possible to prove some results: if RH is false or $\zeta(s)$ has a zero of order at least $2$, then $\sum_{p \leq x} \frac{1}{\sqrt{p}} - \int_{2}^{x} \frac{dt}{\sqrt{t} \log t}$ is unbounded from both above and below. Unconditionally, one can show that this does not converge to zero. Finally, if one assumes RH plus a conjecture called the Linear Independence hypothesis (which is believed to be true by most analytic number theorists), then again it can be shown that this is unbounded from both above and below. – Peter Humphries Mar 26 '17 at 20:52
  • None of these results actually appear anywhere in the literature, but the method is well-known; the general idea is in chapter 15 of Montgomery and Vaughan (or see my paper on weighted sums of the Liouville function). – Peter Humphries Mar 26 '17 at 20:53