I'm supposed to compute the asymptotic density of the set \begin{equation} \Pi(x):=\#\{p^k \leq x :p \;prime, k \in \mathbb{N}\} \end{equation} of prime powers less or equal to $x$, that is, compute $\lim_{x \to \infty} \dfrac {\Pi(x)}{x}$.
So $\Pi(x)$ denotes the number of elements in the set $\{p^k \leq x :p \;prime, k \in \mathbb{N}\}$? I have never seen this notation. I am not looking for a solution here, just for hints.
$\endgroup$ 103 Answers
$\begingroup$Copied from another answer of mine:
The following result is from the paper Ivan Niven. The asymptotic density of sequences. Bull. Amer. Math. Soc., 57(6):420-434, 1951. Often it can be used to show that asymptotic density of some set is $0$.
Theorem 7. If for a set of primes $\{p_i\}$ we have $d(A_{p_i})=0$ for every $i$, and if $\sum p_i^{-1}=\infty$ then $d(A)=0$.
Here, for any $A\subseteq\mathbb N$ and a prime $p$, the set $A_p$ is defined as $$A_p=\{n\in A; p\mid n, p^2\nmid n\}.$$
Note: If we defined $A_p=\{n\in A; p\mid n\}$ instead of the above result, we would get a weaker result, but the proof is much simpler.
Now if you use the above result for $A=\{p^k; k\ge 1\}$ (and take $\{p_i\}$ to be the set of all primes), you easily get that $d(A)=0$.
EDIT: Originally I considered only the set $A=\{p^k; k\ge 2\}$ and discussed the density of primes separately. This is not necessary. (Which is why the following paragraph is not needed anymore.)
So now it remains to find asymptotic density of the set of all primes (since higher powers do not contribute to asymptotic density). Several proofs that $d(\mathbb P)=0$ are in this post: Percentage of primes among the natural numbers
Edit: TonyK's comment (amplifying Antonio Vargas') answers the question. The following is a sketch of an alternative argument.
$\psi(x) = \sum_{p^k \leq x }\log p \sim x$ [PNT]. Using your (or the comments) definition of $\Pi(x)$ if we have, for example, $x = 20,$
$$ \psi(x) = \log 2+ \log 2^2 +\log 2^3 +\log 2^4 +\log 3 + \log 3^2 + \log 5+ ... + ~\log 19.$$
Rewrite as:
$(\log 2\cdot 2^2\cdot 2^3\cdot2^4) + (\log 3\cdot 3^2) +...+~ (\log 19) = (1+2+3+4) \log 2 + (1+2)\log 3 +...,$ etc.
Notice that the cardinality of $p^k$ is the highest power in each of the terms on the left side of the equality above, or the highest term in the coefficients of the logs, $(1+2+3+4),~ (1+2),$ etc. For 20,
$$ \frac{\Pi(20)}{\psi(20)} = \frac{4 + 2+~ ...~ + 1}{10 \log 2 + 3\log 3+~...~+ \log 19} $$
and as x grows the coefficients in the denominator are $k(k+1)/2$ while the corresponding terms in the numerator are k. In the example above, $10 = (4\cdot 5) /2$ and $3 = (2\cdot 3)/2.$ Since $\log n > 1$ for $n \geq 3$ the terms in the denominator are growing a lot faster than those in the numerator so
$$\frac{\Pi(x)}{x} \sim \frac{\Pi(x)}{\psi(x)} \sim 0. $$
Using Mathematica and the formula suggested in the comments above I calculated $\frac{\Pi(x)}{x}$ for x = 10, 100, 1000, 10000, 100000. I get:
0.7, 0.35, 0.193, 0.128, 0.097, so the limit appears to be zero.
$\endgroup$ 5 $\begingroup$You can prove stronger results; in increasing order of generality we have
- The upper Banach density of prime powers is 0.
- The upper Banach density of perfect powers is 0.
- The Buck's measure density of perfect powers is 0.
All these results follow from Corollary 6 in 1.
$\endgroup$
