1

Let $f(x)$ be the number of prime powers less than $x$
(by prime power I mean any number of the form $p^n$ such that $p$ is prime and $n \ge 1$.)

Consider the limit: lim x->∞ f(x)/x
I would like to know the value of L
In particular, this article seems to suggest that the limit tend neither towards 0 nor 1 however I do not have the knowledge needed to understand the it:

http://onlinelibrary.wiley.com/doi/10.1002/cpa.3160270602/abstract

Thanks

Yakov Shklarov
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Mathew
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  • Are you counting the $n = 1$ terms, i.e. the primes themselves, or just their higher powers? – John Barber Jan 24 '18 at 04:55
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    Isn't $f(x) = \sum_{n=1}^{\infty} \Pi\left(x^{1/n}\right)$? – John Barber Jan 24 '18 at 05:03
  • I don't know, if you can prove that or give me a link to a proof I'll accept that as an answer – Mathew Jan 24 '18 at 05:05
  • Use \lim_{x\to\infty}\frac{f(x)}{x} to better express $L$ in symbols. This generates thus: $$L = \lim_{x\to\infty}\frac{f(x)}{x}$$ You can also use \cfrac and \dfrac to change the size of the fraction. – Mr Pie Jan 24 '18 at 05:08
  • I tried it and the result aren't good, I'm probably making a simple mistake, feel free to point it out – Mathew Jan 24 '18 at 05:12
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    That is probably theworst URL in existence. – Mariano Suárez-Álvarez Jan 24 '18 at 05:15
  • sorry about that, if you know how to hide the url in the link let me know and I'll change it or change it your self if you can – Mathew Jan 24 '18 at 05:17
  • Here is a better link to the article referenced above. – John Barber Jan 24 '18 at 05:40
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    This article is not about what you think. It's investigating the distribution of the fractional part of $p^\gamma$ for primes $p$ and some $0<\gamma<1$, and those fractional parts are uniformly distributed in $[0,1]$, not surprisingly. –  Jan 24 '18 at 07:42

2 Answers2

0

Let $N=\lfloor \log_2(x)\rfloor$. Then \begin{align}f(x)-\pi(x)=\sum_{\substack{p^m\le x \\ m\ge 2}}1=&\sum_{p^2\le x}1+\sum_{p^3\le x}1+\cdots +\sum_{p^N\le x}1\\[1mm] =&\sum_{p\le x^{1/2}}1+\sum_{p\le x^{1/3}}1+\cdots +\sum_{p \le x^{1/N}}1\\[1mm] =&\;\pi(x^{1/2})+\pi(x^{1/3})+\cdots +\pi(x^{1/N})\\[1mm] \le&\;(N-1)\pi(x^{1/2})\\[1mm] =&\;O\Big(\frac{\sqrt{x}}{\log x}\log_2(x)\Big)=O(\sqrt{x}) \end{align} Thus, $$\frac{f(x)}{x}=\frac{\pi(x)}{x}+O(\frac{1}{\sqrt{x}})=O(\frac{1}{\log x})=o(1)$$ by the prime number theorem. As such, we have $L=0$.

AfterMath
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-2

Let's define $\Pi_n(x)$ as the number of prime powers $p^n$ with exponent $n$ that are less than $x$. Clearly $\Pi_1(x) = \Pi(x)$, the usual prime counting function.

Since $p^n < x$ implies that $p < x^{1/n}$, we have $$ \Pi_n(x) = \Pi\left(x^{1/n}\right)\, . $$ Your $f(x)$ is: \begin{align} f(x) &= \sum_{n=1}^{\infty} \Pi_n(x)\\ &= \sum_{n=1}^{\infty}\Pi\left(x^{1/n}\right)\, . \end{align} For fixed $x$, $x^{1/n}$ will eventually go to 1 as $n\rightarrow\infty$, so this sum will actually terminate at some point. (Actually the sum need only go up to the point where $x^{1/n} < 2$, i.e. $n_{max} = \left\lfloor \ln x / \ln 2 \right \rfloor$.)

Beyond that, at present, all I have are numerics (done in Mathematica) which loosely suggest that $$ \lim_{x\rightarrow\infty} \frac{f(x)}{x} = 0\, , $$ although convergence seems to be very, very slow. (Perhaps going as $1/\ln(x)$?)

Edit:

Numerically, it looks like the limit $$ \lim_{x\rightarrow\infty} \frac{f(x)}{x/\ln(x)} $$ may exist, and have a value $\ge 1$. (I suspect the actual limit is 1 itself, but the convergence is very slow.)

Edit 2:

For those down-voting this, note that the original questioner asked me (in the comments above) to post the proof of $f(x) = \sum_{n=1}^{\infty}\Pi\left(x^{1/n}\right)$. I'm just trying to be helpful. (Unless I've made an error here, in which case let me know.)

John Barber
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