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

Question

$N$ 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 $n$).

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My approach:

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

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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?

Answer 1

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

Answer 2

To answer the 2nd part to the question, note that the complement of the event $p_i^2 \nmid N$ is $p_i^2 \mid N$. Let $A_i$ denote this event. We know that all primes are coprime, so their squares are as well. Thus, if squares $p_{i_1}^2, ... p_{i_n}^2$ all divide $N$, it must be that $p_{i_1}^2...p_{i_n}^2$ divides $N$. This occurs with probability $1/p_{i_1}^2...p_{i_n}^2$, so

$$P(A_{i_1}...A_{i_n}) = P(A_{i_1})...P(A_{i_n}) = \frac{1}{p_{i_1}^2...p_{i_n}^2}$$

Hence, the events $A_1, ...$ are all independent. Complements of independent events are independent, and that completes the proof.