# Zeta Equivalence to Prime Number Theorem

## Theorem

Let $\map \zeta z$ be the Riemann $\zeta$ function.

The Prime Number Theorem is logically equivalent to the statement that the average of the first $N$ coefficients of $\dfrac {\zeta'} {\zeta}$ tend to $-1$ as $N$ goes to infinity.

## Proof

The Von Mangoldt Equivalence is equivalent (clearly) to the statement that the average of the coefficients of the function of $z$ defined as:

$(1): \quad \displaystyle \sum_{n \mathop = 1}^\infty \frac {\map \Lambda n} {n^z}$

tend to $1$.

Let $\set {p_1, p_2, p_3, \dots}$ be an enumeration of the prime numbers:

$\set { 2, 3, 5, 7, 11, \dots}$

In the proof of the Von Mangoldt Equivalence, in the sum of von Mangoldt function, the $\map \ln p$ term will appear once for each power of $p$.

So, we expand out $(1)$ as:

 $\displaystyle \sum_{n \mathop = 1}^\infty \frac{\map \Lambda n} {n^z}$ $=$ $\displaystyle \map \ln {p_1} \paren {\frac 1 {p_1^z} + \frac 1 {p_1^{2 z} } + \frac 1 {p_1^{3 z} } + \cdots} + \map \ln {p_2} \paren {\frac 1 {p_2^z} + \frac 1 {p_2^{2 z} } + \cdots} + \cdots$ $\displaystyle$ $=$ $\displaystyle \map \ln {p_1} \sum_{n \mathop = 1}^\infty \paren {\paren {p_1^{-z} }^n} + \map \ln {p_2} \sum_{n \mathop = 1}^\infty \paren {\paren {p_2^{-z} }^n} + \cdots$ $\displaystyle$ $=$ $\displaystyle \map \ln {p_1} \frac {p_1^{-z} } {1 - p_1^{-z} } + \map \ln {p_2} \frac {p_2^{-z} } {1 - p_2^{-z} } + \cdots$ Sum of Infinite Geometric Sequence $\displaystyle$ $=$ $\displaystyle \sum_{p \text{ prime} } \map \ln p \frac {p^{-z} } {1 - p^{-z} }$

This function of $z$ can be recognized as:

 $\displaystyle \sum_{p \text{ prime} } \map \ln p \frac {p^{-z} } {1 - p^{-z} }$ $=$ $\displaystyle \sum_{p \text{ prime} } \paren {1 - p^{-z} } \frac {-\paren {0 - \map \ln p p^{-z} } } {\paren {1 - p^{-z} }^2}$ $\displaystyle$ $=$ $\displaystyle \sum_{p \text{ prime} } \frac \d {\d z} \map \ln {\frac {-1} {1 - p^{-z} } }$ $\displaystyle$ $=$ $\displaystyle \map {\frac \d {\d z} } {\sum_{p \text{ prime} } \map \ln {\frac {-1} {1 - p^{-z} } } }$ $\displaystyle$ $=$ $\displaystyle \map {\frac \d {\d z} } {\ln \prod_{p \text{ prime} } \frac {-1} {1 - p^{-z} } }$ $\displaystyle$ $=$ $\displaystyle -\frac \d {\d z} \map \ln {\map \zeta z}$ $\displaystyle \prod_{p \text{ prime} } \frac 1 {1 - p^{-z} }$ is the Riemann zeta function $\displaystyle$ $=$ $\displaystyle -\frac {\map {\zeta'} z} {\map \zeta z}$

Hence the result.

$\blacksquare$