Von Mangoldt Equivalence

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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:

\(\ds \sum_{n \mathop = 1}^N \map \Lambda n\) \(=\) \(\ds \map \Lambda 1 + \map \Lambda 2 + \cdots + \map \Lambda n\)
\(\ds \) \(=\) \(\ds 0 + \map \ln 2 + \map \ln 3 + \map \ln 2 + \map \ln 5 + 0 + \map \ln 7 + \map \ln 2 + \map \ln 3 + 0 + \cdots\)

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.

$\blacksquare$


Source of Name

This entry was named for Hans Carl Friedrich von Mangoldt.