Von Mangoldt Equivalence

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

Then:
 * $\displaystyle \lim_{N \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \Lambda \left({n}\right) = 1$

is logically equivalent to the Prime Number Theorem.

Proof
Observe:

Notice this sum will have:
 * as many $\ln \left({2}\right)$ terms as there are powers of $2$ less than or equal to $N$
 * as many $\ln \left({3}\right)$ terms as there are powers of $3$ less than or equal to $N$

and in general, if $p$ is a prime less than $N$, $\ln \left({p}\right)$ will occur in this sum $\left \lfloor {\log_p \left({N}\right)}\right \rfloor$ times.

But:
 * $\displaystyle \ln \left({p}\right) \left \lfloor {\log_p \left({N}\right)} \right \rfloor \sim \ln \left({p}\right) \log_p \left({N}\right) = \ln \left({N}\right)$

so:
 * $\displaystyle \sum_{p \text{ prime} \mathop \le N} \ln \left({p}\right) \left \lfloor {\log_p \left({N}\right)} \right \rfloor \sim \sum_{p \text{ prime} \mathop \le N} \ln \left({N}\right) = \pi \left({N}\right) \ln \left({N}\right)$

Therefore:
 * $\displaystyle \sum_{n \mathop = 1}^N \Lambda \left({n}\right) \sim \pi \left({N}\right) \ln \left({N}\right)$

and so if:
 * $\displaystyle \lim_{N \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \Lambda \left({n}\right) = 1$

then:
 * $\displaystyle \lim_{N \to \infty} \frac 1 N \pi \left({N}\right) \ln \left({N}\right) = 1$

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.