# Probability that applying Möbius Function to randomly chosen positive integer is non-zero.

$$N$$ is a randomly selected positive integer from the set $$\{1,2,3,\ldots,10^k\}$$. The task is to find $$P\big(\mu(N)\neq0\big)$$ as $$k\rightarrow\infty$$, where $$\mu(n)=0$$ if $$n$$ has a repeated prime factor (i.e. $$\exists\ p$$ such that $$p$$ is prime and $$p^2$$ divides $$n$$).

My approach:

$$\mu(N)\neq0 \iff 2^2\nmid N, \ 3^2\nmid N, \ 5^2\nmid N,\ldots$$ $$\Rightarrow P\big(\mu(N)\neq0\big)=P(2^2\nmid N, \ 3^2\nmid N, \ 5^2\nmid N,\ldots)$$ $$=\prod_{i=1}^\infty P(p_i^2 \nmid N) \ \ \ \ where \ p_i \ is \ the \ i_{th} \ smallest \ prime$$ $$=\prod_{i=1}^\infty 1-\frac1{p_i^2} \ \ \ (\because P(a\mid N)\rightarrow \frac1{a} when \ k\rightarrow \infty)$$ $$=\prod_{i=1}^\infty \frac{p_i^2-1}{p_i^2}=\frac{6}{\pi^2}$$

My question is two-part:

$$1)$$ The book I'm using states $$\prod_{i=1}^\infty \frac{p_i^2-1}{p_i^2}=\frac{6}{\pi^2}$$ as a hint, but I am unsure of where this even comes from. Could someone explain why this holds?

$$2)$$ This approach seems a little shaky to me because I have assumed independence of the events $$2^2\nmid N, 3^2\nmid N,\ldots$$ to evaluate the probability of their intersection as a product of their individual probabilities. How do I justify this independence?

The identity follows from using the Euler product of $$\zeta(2) = \pi^2/6$$. Proving this equality involves the Fourier series of $$f(x)=x$$.