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 

$$\sum_{n=1}^{\infty} \frac{\mu(n)}{n}=o(1),$$
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.