Prime Number Theorem

Theorem
The prime-counting function $\pi \left({n}\right)$, that is, the number of primes less than $n$, satisfies:
 * $\displaystyle \lim_{n \to \infty} \pi \left({n}\right) \frac {\ln \left({n}\right)} n = 1$

or equivalently,
 * $\displaystyle \pi \left({n}\right) \sim \frac n {\ln \left({n}\right)}$

Proof
The proof presented here is a version of Donald Newman's proof. For ease of reading, the proof is broken into parts, with the goal of each part presented.

From the Von Mangoldt Equivalence, the Prime Number Theorem is logically equivalent to:


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

where $\Lambda$ is the von Mangoldt function.

While useful, the von Mangoldt function is a discrete function that is not very much easier to work with than $\pi \left({n}\right)$ itself.

It behooves us to find another statement equivalent to the Prime Number Theorem.

From Zeta Equivalence to Prime Number Theorem, 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.

Now we demonstrate the truth of this claim regarding $\dfrac{\zeta'}{\zeta}$.

Doing so proves the Prime Number Theorem.

We know that all of the coefficients of $\zeta$ are precisely $1$.

So the statement:
 * The average of the first $N$ coefficients of $\frac{\zeta'}{\zeta}$ tend to $1$ as $N$ goes to infinity

is equivalent to the statement:
 * The average of the first $N$ coefficients of $\frac{\zeta'}{\zeta} - \zeta$ tend to $0$ as $N$ goes to infinity.

The latter will be more convenient for our purposes.

We write:


 * $\displaystyle \frac{\zeta'(z)}{\zeta(z)} - \zeta(z) = \frac{1}{\zeta(z)} \left({ \zeta'(z) - \zeta^2 (z) }\right)$

so that we can use knowledge of the inverse and square of $\zeta$, and knowledge of its derivative:


 * $\displaystyle \frac{1}{\zeta(z)} \left({ \zeta'(z) - \zeta^2 (z) }\right) = \left({ \sum_{n \mathop = 1}^\infty \frac{\mu \left({n}\right)}{n^z} }\right)

\left({ \left({ \sum_{n \mathop = 1}^\infty \frac{\log \left({n}\right)}{n^z} }\right) - \left({ \sum_{n \mathop = 1}^\infty \frac{d \left({n}\right)}{n^z} }\right) }\right)$

where $\mu \left({n}\right)$ is the Möbius function and $d \left({n}\right)$ is the divisor function.

Given this form of the function, we can see that the average of the first $N$ coefficients is:
 * $\displaystyle \frac 1 N \sum_{ab \mathop \le N} \left({ \mu(a) \left({ \log(b) - d(b) }\right) }\right)$

Hence the Prime Number Theorem is equivalent to the statement that this expression tends to $0$ as $N \to \infty$.

At this point, we can add $0 = 2 \gamma / N - 2 \gamma / N$, where $\gamma$ is the Euler-Mascheroni constant:


 * $\displaystyle = \frac 1 N \sum_{a b \mathop \le N} \left({ \mu(a) \left({ \ln(b) - d(b) }\right) }\right) + 1 \frac {2 \gamma} N - \frac {2 \gamma} N$

Using a convenient property of the Möbius function, this $1$ is just


 * $\displaystyle 1 = \underbrace{ \sum_{a \mathop \backslash 1} \mu(a)}_{=1} + \underbrace{\sum_{a \mathop \backslash 2} \mu(a)}_{=0} + \dots + \underbrace{\sum_{a \backslash N} \mu(a)}_{=0}$


 * $\displaystyle = \sum_{n \mathop = 1}^N \left({ \sum_{a \mathop \backslash n} \mu(a) }\right)$

and so:


 * $\displaystyle = \frac 1 N \sum_{a b \mathop \le N} \left({ \mu(a) \left({ \ln(b) - d(b) }\right) }\right) + 1 \frac{2 \gamma} N - \frac{2 \gamma} N = \frac 1 N \sum_{a b \mathop \le N} \left({ \mu(a) \left({ \ln(b) - d(b) }\right) }\right) + \frac 1 N \sum_{n \mathop = 1}^N \left({ \sum_{a \mathop \backslash n} \mu(a) 2 \gamma }\right) - \frac {2 \gamma} N$


 * $\displaystyle = \frac 1 N \sum_{ab \mathop \le N} \left({ \mu(a) \left({ \ln(b) - d(b) + 2 \gamma }\right) }\right) - \frac {2 \gamma} N$

By Order of Divisor Function:


 * $\displaystyle = \frac 1 N \sum_{a \mathop \le N} \mu(a) O \left({-\sqrt N}\right) - \frac {2 \gamma} N$

and by Order of Möbius Function:


 * $\displaystyle = \frac 1 N o \left({N}\right) O \left({-\sqrt N}\right) - \frac {2 \gamma} N = O \left({\frac{-1}{\sqrt N}}\right) o \left({N}\right) - \frac {2 \gamma} N$

As $N \to \infty$, we have


 * $\displaystyle \lim_{N \to \infty} \left({ O \left({ \frac{-1}{\sqrt N}}\right) o \left({N}\right) - \frac {2 \gamma} N}\right)$

which clearly goes to $0$ as $O \left({\dfrac{-1}{\sqrt N}}\right)$ dominates $o \left({N}\right)$.