Why shouldn't this prove the Prime Number Theorem? 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 ?
 A: 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.
In particular, as pointed out by @TerryTao in the comments:

if $t\neq 0$ is real, then
$$ \lim_{s\rightarrow 1^+} \sum_{n=1}^{\infty} \frac{n^{it}}{n^s},$$
can be shown to converge to a finite value, whereas
  $$\lim_{x\rightarrow \infty} \sum_{n\leq x} \frac{n^{it}}{n}$$
is undefined. So at a bare minimum one has to somehow stop $\mu(n)$ from "pretending" to be like $n^{it}$. This turns out to be basically equivalent to preventing $\zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the PNT.

