distribution of prime powers

Question

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

Answer 1

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.)

Answer 2

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 $L=0$.