Theorem: Prime Number Theorem

The prime-counting function $\map \pi n$, that is, the number of primes less than $n$, satisfies:
:$\ds \lim_{n \mathop \to \infty} \map \pi n \frac {\map \ln n} n = 1$
or equivalently: 
:$\map \pi n \sim \dfrac n {\map \ln n}$
where $\sim$ denotes asymptotic equivalence.

Proof:
{{questionable|The bounds obtained are too tight; the Landau notation calculation does not work}}
The proof presented here is a version of {{AuthorRef|Donald J. 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:
:$\ds \lim_{N \mathop \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \map \Lambda n = 1$
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 $\map \pi n$ 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 $1$.
{{explain|do we?}}
So the statement:
:The average of the first $N$ coefficients of $\dfrac {\zeta'} {\zeta}$ 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:
:$\dfrac {\map {\zeta'} z} {\map \zeta z} - \map \zeta z = \dfrac 1 {\map \zeta z} \paren {\map {\zeta'} z - \map {\zeta^2} z}$
From Reciprocal of Riemann Zeta Function, Square of Riemann Zeta Function and Derivative of Riemann Zeta Function:
:$\ds \frac 1 {\map \zeta z} \paren {\map {\zeta'} z - \map {\zeta^2} z} = \paren {\sum_{n \mathop = 1}^\infty \frac {\map \mu n} {n^z} } \paren {\paren {\sum_{n \mathop = 1}^\infty \frac {\map \ln n} {n^z} } - \paren {\sum_{n \mathop = 1}^\infty \frac {\map {\sigma_0} n} {n^z} } }$
where:
:$\map \mu n$ is the Möbius function
:$\map {\sigma_0} n$ is the divisor count function.
Given this form of the function, we can see that the average of the first $N$ coefficients is:
:$\ds \frac 1 N \sum_{a b \mathop \le N} \paren {\map \mu a \paren {\map \ln b - \map {\sigma_0} b} }$
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:
{{begin-eqn}}
{{eqn | l = 0
      | r = \dfrac {2 \gamma} N - \dfrac {2 \gamma} N
      | c = where $\gamma$ is the Euler-Mascheroni constant
}}
{{eqn | r = \frac 1 N \sum_{a b \mathop \le N} \paren {\map \mu a \paren {\map \ln b - \map {\sigma_0} b} } + 1 \frac {2 \gamma} N - \frac {2 \gamma} N
      | c = 
}}
{{end-eqn}}
{{explain|It's not sure what we are trying to do here. We seem to be assuming what we want to prove.}} 
From Sum of Möbius Function over Divisors, this $1$ is just:
:$\ds 1 = \underbrace {\sum_{a \mathop \divides 1} \map \mu a}_{= 1} + \underbrace {\sum_{a \mathop \divides 2} \map \mu a}_{= 0} + \dots + \underbrace {\sum_{a \mathop \divides N} \map \mu a}_{= 0}$
Hence we continue from the above:
{{begin-eqn}}
{{eqn | l = 0
      | r = \frac 1 N \sum_{a b \mathop \le N} \paren {\map \mu a \paren {\map \ln b - \map {\sigma_0} b} } + 1 \frac {2 \gamma} N - \frac {2 \gamma} N
      | c = 
}}
{{eqn | r = \frac 1 N \sum_{a b \mathop \le N} \paren {\map \mu a \paren {\map \ln b - \map {\sigma_0} b} } + \frac 1 N \sum_{n \mathop = 1}^N \paren {\sum_{a \mathop \divides n} \map \mu a 2 \gamma} - \frac {2 \gamma} N
      | c = 
}}
{{eqn | r = \frac 1 N \sum_{a b \mathop \le N} \paren {\map \mu a \paren {\map \ln b - \map {\sigma_0} b + 2 \gamma} } - \frac {2 \gamma} N
      | c = 
}}
{{eqn | r = \frac 1 N \sum_{a \mathop \le N} \map \mu a \map \OO {-\sqrt N} - \frac {2 \gamma} N
      | c = Order of Divisor Count Function
}}
{{eqn | r = \frac 1 N \map o N \map \OO {-\sqrt N} - \frac {2 \gamma} N
      | c = 
}}
{{eqn | r = \map \OO {\frac {-1} {\sqrt N} } \map o N - \frac {2 \gamma} N
      | c = Order of Möbius Function
}}
{{end-eqn}}
As $N \to \infty$, we have:
:$\ds \lim_{N \mathop \to \infty} \paren {\map \OO {\frac {-1} {\sqrt N} } \map o N - \frac {2 \gamma} N}$
which clearly goes to $0$ as $\map \OO {\dfrac {-1} {\sqrt N} }$ dominates $\map o N$.
{{explain|More detail needed in the above.}}