I will present a very short proof of the Prime Number Theorem.
My question is, if the following proof is acceptable?
Let ф(np) be the Euler ф function (Euler totient function) for any primorial np with (1) $$np=\prod_{p=prime}^{p≤n} p$$ which is defined as (2) $$ф(np)=np*\prod_{p=prime}^{p≤n} (1-1/p)=np*\prod_{p=prime}^{p≤n} \frac{(p-1)}{p}$$ and where the product extends over all primes p dividing the primorial np.
This function gives the number of all possible primes up to any primorial np, which are not divided by all the primes less than n and all these possible primes between the interval [1,np] are quite uniformly distributed, especially for $\lim_{n\to ∞}$.
We further know that all the possible primes between the interval $[n,n^2]$ are identical with the actual primes.
For $\lim_{n\to ∞}$the primorial np is defined by (3) $$np=e^n$$ and therefore (4): $$\frac{ф(e^n)}{e^n}=\prod_{p=prime}^{∞} \frac{(p-1)}{p}$$
As the possible primes between the interval $[1,np]$ are quite uniformly distributed and because all the possible primes between the interval $[n,n^2]$ are identical with the actual primes, the equation from above also states the prime number density function (5) $$\frac{π(n)}{n}$$ and we can write (6) $$\frac{ф(e^n)}{e^n}=\prod_{p=prime}^{∞} \frac{(p-1)}{p}=\frac{π(n)}{n}$$
From Euler (1737) we also know that for $\lim_{n\to ∞}$ (7)
$$\sum_{n=1}^∞ \frac{1}{n}=\prod_{p=prime}^∞ \frac{p}{(p-1)}$$
and with
$$ln(n)+γ=\sum_{n=1}^∞ \frac{1}{n}$$
and with the Euler-Mascheroni constant γ=0,57721… the two equations from above state that for $\lim_{n\to ∞}$
$$ln(n)≅\sum_{n=1}^∞ \frac{1}{n}=\prod_{p=prime}^∞ \frac{p}{(p-1)}$$
and therefore (8)
$$\frac{1}{ln(n)}≅\prod_{p=prime}^∞ \frac{(p-1)}{p}$$
Equation (6) in conjunction with equation (8) gives (9)
$$\frac{ф(e^n)}{e^n}=\prod_{p=prime}^{∞} \frac{(p-1)}{p}=\frac{π(n)}{n}≅\frac{1}{ln(n)}$$
and therefore (10) for $\lim_{n\to ∞}$
$$\frac{π(n)}{n}≅\frac{1}{ln(n)}$$
and (11)
$$π(n)≅\frac{n}{ln(n)}$$
respectively.
