{{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: :$\displaystyle \lim_{N \mathop \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \map \Lambda n = 1$ 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: :$\displaystyle \frac 1 {\map \zeta z} \paren {\map {\zeta'} z - \map {\zeta^2} z} = \paren {\sum_{n \mathop = 1}^\infty \frac {\map \mu n} {n^z} } \paren {\paren {\sum_{n \mathop = 1}^\infty \frac {\map \ln n} {n^z} } - \paren {\sum_{n \mathop = 1}^\infty \frac {\map d n} {n^z} } }$ where: :$\map \mu n$ is the Möbius function :$\map d n$ is the divisor function. Given this form of the function, we can see that the average of the first $N$ coefficients is: :$\displaystyle \frac 1 N \sum_{a b \mathop \le N} \paren {\map \mu a \paren {\map \ln b - \map d b} }$ 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: :$0 = 2 \gamma / N - 2 \gamma / N$ where $\gamma$ is the Euler-Mascheroni constant: {{tidy|Put the below into eqn template}} :$\displaystyle = \frac 1 N \sum_{a b \mathop \le N} \paren {\map \mu a \paren {\map \ln b - \map d b} } + 1 \frac {2 \gamma} N - \frac {2 \gamma} N$ Using a convenient property of the Möbius function, this $1$ is just: :$\displaystyle 1 = \underbrace {\sum_{a \mathop \divides 1} \map \mu a}_{= 1} + \underbrace {\sum_{a \mathop \divides 2} \map \mu a}_{= 0} + \dots + \underbrace {\sum_{a \mathop \divides N} \map \mu a}_{= 0}$ :$\displaystyle = \sum_{n \mathop = 1}^N \paren {\sum_{a \mathop \divides n} \map \mu a}$ and so: :$\displaystyle = \frac 1 N \sum_{a b \mathop \le N} \paren {\map \mu a \paren {\map \ln b - \map d b} } + 1 \frac {2 \gamma} N - \frac {2 \gamma} N = \frac 1 N \sum_{a b \mathop \le N} \paren {\map \mu a \paren {\map \ln b - \map d b} } + \frac 1 N \sum_{n \mathop = 1}^N \paren {\sum_{a \mathop \divides n} \map \mu a 2 \gamma} - \frac {2 \gamma} N$ :$\displaystyle = \frac 1 N \sum_{a b \mathop \le N} \paren {\map \mu a \paren {\map \ln b - \map d b + 2 \gamma} } - \frac {2 \gamma} N$ By Order of Divisor Function:   :$\displaystyle = \frac 1 N \sum_{a \mathop \le N} \map \mu a \, \map O {-\sqrt N} - \frac {2 \gamma} N$ and by Order of Möbius Function: :$\displaystyle = \frac 1 N \, \map o N \, \map O {-\sqrt N} - \frac {2 \gamma} N = \map O {\frac {-1} {\sqrt N} } \, \map o N - \frac {2 \gamma} N$ As $N \to \infty$, we have  :$\displaystyle \lim_{N \mathop \to \infty} \paren{\map O {\frac {-1} {\sqrt N} } \, \map o N - \frac {2 \gamma} N}$ which clearly goes to $0$ as $\map O {\dfrac {-1} {\sqrt N} }$ dominates $\map o N$. {{explain|More detail needed in the above.}}