Question: 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.

Answer: 
Copied from another [answer of mine](https://math.stackexchange.com/a/126795/):

The following result is from the paper
Ivan Niven. [The asymptotic density of sequences](http://projecteuclid.org/euclid.bams/1183516304). 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](https://math.stackexchange.com/questions/82074/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.