The asymptotic number of
[square-free numbers](http://en.wikipedia.org/wiki/Square-free_integer#Distribution)
$\le n$ is $Q(n) = 6n/\pi^2 + O(\sqrt{n})$.
Because
[$\zeta(2)=\pi^2/6$](http://mathworld.wolfram.com/RiemannZetaFunctionZeta2.html),
$Q(n) \approx n/\zeta(2)$.


[OEIS A004709](http://oeis.org/A004709)
says that cube-free numbers have asymptotic density of
$1/\zeta(3)$, "the reciprocal of
[Apery's constant](http://mathworld.wolfram.com/AperysConstant.html)"
("the probability that three randomly chosen integers are relatively prime":
[link here](http://oeis.org/A088453)).



> 
> ***Q***. Are there analogous results or conjectures for $k^{\textrm{th}}$-power-free numbers?, $k>3$?
>  Does the density continue to $1/\zeta(k)$? Is that conjectured or proven or disproven?
> 
> 
> 


I ask this in (obvious) number-theoretic naiveté.


---


**Answered** by Gjergji Zaimi and Douglas Zare and Noam Elkies: the density indeed grows
as $1/\zeta(k)$, and this has been established.



Square-free grows as $6n/\pi^2$: $k$-th free?
Yes, it works in much the same way for any $k$. Here's an elementary proof.


Let $Q\_k(n)$ be the number of $k$-th power free integers $\leq n$.
Then
$$
Q\_k(n) = \sum\_{d^k \leq n} \mu(d) \lfloor n/d^k \rfloor
= \sum\_{d^k \leq n} \mu(d) \, (n/d^k + \theta\_d)
$$
for some $\theta\_d \in [0,1)$. Hence 
$$
\Bigl| \, Q\_k(n) - \sum\_{d^k \leq n} \frac{\mu(d)}{d^k} n \, \Bigr| < n^{1/k}.
$$
But $\sum\_{d^k \leq n} \mu(d)/d^k$ is a partial sum of
a series that converges to $1/\zeta(k)$, with error bounded by
$\sum\_{d^k > n} 1/d^k \ll n^{1/k}/n$. Therefore $Q\_k(n) = n/\zeta(k) + O(n^{1/k})$,
**QED**.


