# simply proof of prime number theorem

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.

• Calling the $n$'th primorial $np$ makes most of this very hard to read, as that should never mean anything but the product of those numbers (you even have both $n$ and $p$ be numbers, so that product would make sense). – Tobias Kildetoft Sep 26 '17 at 9:11
• "As the possible primes........are quite uniformly distributed". Are you sure? – Konstantinos Gaitanas Sep 26 '17 at 9:35
• Come on $\ln n = \sum_{n=1}^\infty \frac{1}{n}= \prod_p (1-\frac{1}{p})$ ? If you want to see why the prime number theorem is not elementary, take a look at the Mertens theorems. – reuns Sep 26 '17 at 11:38