Ttykuu

Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ can be interpreted as the probability that a randomly chosen integer is $k$-free.

Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form

$$sum_n=1^infty fracmu(n)n=0,$$

since the probability that an integer is ``$1$-free'' is zero ?

You ask:

Denote by $mu$ the Mobius function. It is known that for every integer $k>1$, the number $sum_n=1^infty fracmu(n)n^k$ can be interpreted as the probability that a randomly chosen integer is $k$-free.

Letting $krightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form

$$sum_n=1^infty fracmu(n)n=0,$$

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

$$sum_n=1^infty fracmu(n)n=o(1),$$
and it is relatively easy to prove that

$$lim_srightarrow 1^+ sum_n=1^infty fracmu(n)n^s=0.$$
The real difficulty lies in proving that

$$lim_xrightarrow infty sum_nleq x fracmu(n)n=
lim_srightarrow 1^+ sum_n=1^infty fracmu(n)n^s,$$
which is highly nontrivial and requires intricate arguments.