TITLE: Square-free grows as $6n/\pi^2$: $k$-th free?
QUESTION [5 upvotes]: The asymptotic number of
square-free numbers
$\le n$ is $Q(n) = 6n/\pi^2 + O(\sqrt{n})$.
Because
$\zeta(2)=\pi^2/6$,
$Q(n) \approx n/\zeta(2)$.
OEIS A004709
says that cube-free numbers have asymptotic density of
$1/\zeta(3)$, "the reciprocal of
Apery's constant"
("the probability that three randomly chosen integers are relatively prime":
link here).

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.

REPLY [9 votes]: 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.