You ask:
Denote by $\mu$ the Mobius function. It is known that for every integer $k>1$, the number $\sum_{n=1}^{\infty} \frac{\mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.
Letting $k\rightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form
$$\sum_{n=1}^{\infty} \frac{\mu(n)}{n}=0,$$
since the probability that an integer is ``$1$-free'' is zero ?
As pointed out by the users @wojowu and @PeterHumphries, it is true that the PNT is equivalent to
$$\lim_{x \to \infty} \sum_{n\leq x} \frac{\mu(n)}{n}=0,$$ and it is relatively easy to prove that
$$\lim_{s\rightarrow 1^+} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}=0.$$ The real difficulty lies in proving that
$$\lim_{x\rightarrow \infty} \sum_{n\leq x} \frac{\mu(n)}{n}= \lim_{s\rightarrow 1^+} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s},$$ which is highly nontrivial and requires intricate arguments.
