Von Mangoldt Equivalence

Theorem
For $n \in \N_{>0}$, let $\map \Lambda n$ be the von Mangoldt function.

Then:
 * $\ds \lim_{N \mathop \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \map \Lambda n = 1$

is logically equivalent to the Prime Number Theorem.

Proof
Observe:

Notice this sum will have:
 * as many $\map \ln 2$ terms as there are powers of $2$ less than or equal to $N$
 * as many $\map \ln 3$ terms as there are powers of $3$ less than or equal to $N$

and in general, if $p$ is a prime less than $N$, $\map \ln p$ will occur in this sum $\floor {\map {\log_p} N}$ times.

But:
 * $\map \ln p \floor {\map {\log_p} N} \sim \map \ln p \map {\log_p} N = \map \ln N$

so:
 * $\ds \sum_{p \text{ prime} \mathop \le N} \map \ln p \floor {\map {\log_p} N} \sim \sum_{p \text{ prime} \mathop \le N} \map \ln N = \map \pi N \map \ln N$

Therefore:
 * $\ds \sum_{n \mathop = 1}^N \map \Lambda n \sim \map \pi N \map \ln N$

and so if:
 * $\ds \lim_{N \mathop \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \map \Lambda n = 1$

then:
 * $\ds \lim_{N \mathop \to \infty} \frac 1 N \map \pi N \map \ln N = 1$

and vice versa.

But this last equation is precisely the Prime Number Theorem.

Hence our statement regarding the von Mangoldt function is logically equivalent to the Prime Number Theorem.