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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.

Martin Sleziak
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  • I've edited the title to be a little more descriptive. Cheers – Antonio Vargas Jul 08 '14 at 03:38
  • @AntonioVargas, I am trying to prove that it equals 1, do you think this is reasonable? –  Jul 08 '14 at 03:49
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    It's not really my field so I can't say for sure, but if I understand this and this correctly then you should have $\Pi(x) \sim \pi(x)$ and hence $\Pi(x)/x \to 0$ as $x \to \infty$. I'd wait for someone more familiar with these functions to chime in though. – Antonio Vargas Jul 08 '14 at 04:27
  • In one of your links, it says $\Pi(x)= \sum_{n \leq x}\dfrac {\Lambda(n)}{\log n}$, is this really true? –  Jul 08 '14 at 05:02
  • Under the Wiki entry for Prime Number Theorem, Chebyshev's $\psi(x) = \sum_{p^k \leq x} \log p$ is asymptotic to x. Comparing this to your $\Pi(x)$ would seem to confirm Antonio Vargas' answer and I think give a proof. You should write $\psi(x)$ as a product. – daniel Jul 08 '14 at 05:30
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    @daniel Thank you, I am having some troubles with clearly understanding $\Pi(x)$ as I defined it above (its from a homework), does the function give us the number of elements in the set? Like if $p_1 \leq x$ is a prime then we consider the powers of $p_1$ until we arrive at $p_m > x$ for some $m$ and then $m-1$ elements are counted to the value of the function. Is this correct or am I missing something? –  Jul 08 '14 at 05:32
  • That's how I understand it although I am not familiar with $\Pi(x)$ in that form. It is the cardinality of the set of powers of primes less than or equal to x. At least that's how it looks to me. – daniel Jul 08 '14 at 05:34
  • $p^k \le x \iff p \le x^{1/k}$, so $\Pi(x)$ is just $\sum_{k \ge 1}\pi(x^{1/k})$. With a bit of work, you can use this to show that your limit is indeed $1$. – TonyK Jul 08 '14 at 12:13
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    @daniel: Sorry, you are absolutely right. I meant to say that $\Pi(x)/\pi(x) \to 1$. – TonyK Jul 08 '14 at 22:11
  • See also (duplicate?): http://math.stackexchange.com/questions/288439/is-the-asymptotic-density-of-powers-of-primes-zero – Martin Sleziak Jul 09 '14 at 08:07

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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

Community
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Martin Sleziak
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    I have to admit that the proof given in this answer is much simpler. Nevertheless, the result which I mention in my answer is quite often useful, so it is not bad thing to know about it. – Martin Sleziak Jul 09 '14 at 08:14
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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.

daniel
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  • Correct me if I am wrong, it is late over here but cant we just write: $\Pi(x) \leq \pi(x)$? Equality follows if for every $p \leq x$ we have $p^2 > x$, and then since $\pi(x)/x \to 0$ as $x \to \infty$, we are done. –  Jul 08 '14 at 06:06
  • Thanks for your help though, good night :) –  Jul 08 '14 at 06:12
  • Well--not so sure. $\pi(11) = 5.$ But we have $2,2^2,2^3,3,3^2,5,7,11$ or cardinality of 8...? – daniel Jul 08 '14 at 06:14
  • Haha, the inequality is of course the other way around, $\pi(x) \leq \Pi(x)$!!! Missed that...this is bad. –  Jul 08 '14 at 06:22
  • So it's less than $1$ - isn't that obvious??? – barak manos Jul 08 '14 at 12:58
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You can prove stronger results; in increasing order of generality we have

  1. The upper Banach density of prime powers is 0.
  2. The upper Banach density of perfect powers is 0.
  3. The Buck's measure density of perfect powers is 0.

All these results follow from Corollary 6 in 1.

Paolo Leonetti
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