Why does $\pi$ appear in the probability of a number being square-free? The probability of a number being square-free (i.e., the number has no divisor that is a square, cube, etc.) is $6/\pi^2$. I have seen many appearances of $\pi$, and this is also similar to them. All of them can be explained intuitively why pi appears in them. But I don't see any connection between square-free numbers and $\pi$. So my question is:

Why $\pi$ appears here?

Note: I don't want rigorous proofs, I want intuitive explanations.
 A: The most intuitive thing I can think of would be the Riemann Zeta function. You are working with squares, and you will be adding a bunch of squares. In a similar way, if you wanted cube free numbers, you would get Apéry's constant.
A: *

*$6/\pi^2$ appears in the density of square-free numbers because  $\zeta(2)=\pi^2/6$ and every integer is uniquely of the form $k n^2$ with $k$ square-free.


*Then $\zeta(2)=\pi^2/6$ because $$\frac{\pi^2}{\sin^2(\pi z)} = \sum_n \frac1{(z-n)^2}$$
This latter formula is magic: the LHS minus the RHS is a bounded entire function, thus constant, which is the great achievement of complex analysis.


*If you don't like it then the video mentioned in the comment is saying that $$\sum_{n=0}^{2^k} \frac1{|2^k (e^{2i \pi (2n+1)/2^{k+1}}-1)|^2}=1/4$$ and that $$\lim_{k\to \infty} \sum_{n=0}^{2^k} \frac1{|2^k (e^{2i \pi (2n+1)/2^{k+1}}-1)|^2}=\sum_{n=-\infty}^\infty \lim_{k\to \infty}  \frac1{|2^k (e^{2i \pi (2n+1)/2^{k+1}}-1)|^2}$$ $$ = \sum_{n=-\infty}^\infty \frac1{ |i\pi (2n+1)|^2} = \frac1{\pi^2}2(1-1/4)\zeta(2)$$
