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 ?



Why shouldn't this prove the Prime Number Theorem?
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.
> 
> 
> 


