# Zeta Equivalence to Prime Number Theorem

## Theorem

Let $\zeta \left({z}\right)$ 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:

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

tend to $1$.

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

$\left\{{ 2, 3, 5, 7, 11, \dots }\right\}$

In the proof of the Von Mangoldt Equivalence, in the sum of von Mangoldt function, the $\ln \left({p}\right)$ term will appear once for each power of $p$.

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

 $\displaystyle \sum_{n \mathop = 1}^\infty \frac{\Lambda \left({n}\right)}{n^z}$ $=$ $\displaystyle \ln \left({p_1}\right) \left({\frac 1 {p_1^z} + \frac 1 {p_1^{2z} } + \frac 1 {p_1^{3z} } + \cdots}\right) + \ln \left({p_2}\right) \left({\frac 1 {p_2^z} + \frac 1 {p_2^{2z} } + \cdots}\right) + \cdots$ $\displaystyle$ $=$ $\displaystyle \ln \left({p_1}\right) \sum_{n \mathop = 1}^\infty \left({\left({ p_1^{-z} }\right)^n}\right) + \ln \left({p_2}\right) \sum_{n \mathop = 1}^\infty \left({\left({p_2^{-z} }\right)^n}\right) + \cdots$ $\displaystyle$ $=$ $\displaystyle \ln \left({p_1}\right) \frac{p_1^{-z} }{1 - p_1^{-z} } + \ln \left({p_2}\right) \frac{p_2^{-z} }{1 - p_2^{-z} } + \cdots$ Sum of Infinite Geometric Progression $\displaystyle$ $=$ $\displaystyle \sum_{p \text{ prime} } \ln \left({p}\right) \frac{p^{-z} }{1 - p^{-z} }$

This function of $z$ can be recognized as:

 $\displaystyle \sum_{p \text{ prime} } \ln \left({p}\right) \frac{p^{-z} }{1 - p^{-z} }$ $=$ $\displaystyle \sum_{p \text{ prime} } \left({1 - p^{-z} }\right) \frac{-\left({0 - \ln \left({p}\right) p^{-z} }\right)} {\left({1 - p^{-z} }\right)^2}$ $\displaystyle$ $=$ $\displaystyle \sum_{p \text{ prime} } \frac {\mathrm d}{\mathrm dz} \ln \left({\frac {-1} {1 - p^{-z} } }\right)$ $\displaystyle$ $=$ $\displaystyle \frac {\mathrm d}{\mathrm dz} \left({\sum_{p \text{ prime} } \ln \left({ \frac{-1}{1-p^{-z} } }\right)}\right)$ $\displaystyle$ $=$ $\displaystyle \frac {\mathrm d}{\mathrm dz} \left({\ln \prod_{p \text{ prime} } \frac{-1}{1-p^{-z} } }\right)$ $\displaystyle$ $=$ $\displaystyle -\frac {\mathrm d}{\mathrm dz} \ln \left({\zeta \left({z}\right)}\right)$ $\displaystyle \prod_{p \text{ prime} } \frac 1 {1 - p^{-z} }$ is the Riemann zeta function $\displaystyle$ $=$ $\displaystyle -\frac {\zeta' \left({z}\right)} {\zeta \left({z}\right) }$

Hence the result.

$\blacksquare$