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 ?