## The Three Primes Theorem

A new proof of a theorem of Erdos and Turan.
Author: MARTIN MAULHARDT

ABSTRACT. A sequence of three consecutive primes is said to have the closer property if the difference between the first two of them is greater or equal than the difference between the second two. The first three such terns are (3,5,7), (7,11,13) and (13, 17, 19). In this paper we prove using only divergent series that there are infinitely many such terns of primes. This result was first found by P. Erdos and P. Turan in their 1948 paper "On some new questions on the distribution of prime numbers". But they used a quite advanced inequality about primes, namely that π(x) > c x / ln(x). We now conclude this result once again by completely different and simpler arguments and incidentally we find a new criteria of convergence of series.

INTRODUCTION

We first define the main concept of our work.

Definition 1: Let $p_1 < p_2 < p_3 <$ ... $< p_n<$ ... be the sequence of prime numbers.
Let $p_k$, $p_{k+1}$, $p_{k+2}$ be three consecutive terms of that sequence. Let the difference between the first two of them be: $d_k=p_{k+1}-p_k$ and the difference between the two second ones be $d_{k+1}=p_{k+2}-p_{k+1}$.

We say that the tern ( $p_k, p_{k+1}, p_{k+2}$) has the closer property if $d_{k+1} \leq d_k$.

Three examples are (3,5,7), (7,11,13) and (13, 17, 19). In the first of these special cases we have $d_{k+1}=d_k=2$. In the second one we have $d_{k+1}=2$ and $d_k=4$. And finally in the third example we have $d_{k+1}=2$ and $d_k=4$.

A question naturally arises:
Are there infinitelly many such terns in the sequence of primes?

The answer is in the affirmative as we will prove in the next paragraphs.

We first need a new theorem on convergent series.

A NEW CRITERIA OF CONVERGENCE

Theorem 1.
Let $a_1,a_2,a_3,$ ..., $a_n,$ ... be any sequence of natural numbers ordered in increasing magnitude. Let the consecutive first difference between them be $d_n=a_{n+1}-a_n$ for n=1,2,3 .... If this differences form an increasing sequence then the series $\sum\limits_{n=1}^{\infty}\frac{1}{a_n}$

is convergent.

Proof
We have by definition of $d_k$: $d_1 = a_2 - a_1$ $d_2 = a_3 - a_2$ $d_3 = a_4 - a_3$

... $d_n = a_{n+1} - a_n$

Because the $d_i$ are natural numbers and because they're increasing in magnitude we deduce the inequalities $a_2=a_1+d_1\geq a_1+1$ $a_3=a_2+d_2 = a_1+d_1+d_2 \geq a_1 + 1 + 2$

... $a_n\geq a_1+1+2+3$ + … + (n-1) = $a_1+\frac {n(n-1)}{2}$

Taking reciprocals and summing over n, we obtain: $\sum\limits_{n=1}^{\infty}\frac{1}{a_n}$ = $\frac {1}{a_1} + \sum\limits_{n=2}^{\infty}\frac{1}{a_n} \leq 1+ \sum\limits_{n=2}^{\infty}\frac{1}{a_1+\frac {n(n-1)}{2}} \leq$ $\leq 1+ \sum\limits_{n=2}^{\infty}\frac{1}{\frac {n(n-1)}{2}} = 1+ 2 \sum\limits_{n=2}^{\infty}\frac{1}{n(n-1)}<\infty$

which is the desired result.

Examples. A trivial example is the sequence 1,2,4,8,..., $2^n$ ....The consecutive differences are precisely the same numbers 1,2,4,8, ..., $2^n$ ... which form an increasing sequence. Then the trivial result that the series $\sum\limits_{n=1}^{\infty}\frac{1}{2^n}$ is convergent.

A less trivial example is the sequence $F_n$ of the Fibonacci numbers. This sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, ... and the difference between the consecutive terms is the similar sequence 0, 1, 2, 2, 3, 5, 8, 13 ....
This differences form an increasing sequence, then by application of Theorem 1 $\sum\limits_{n=1}^{\infty}\frac{1}{F_n} <\infty$.

Remark. Theorem 1 remains true if we replace the condition that the difference between the two terms form an increasing sequence by the weaker condition that they form an increasing sequence starting at some point $n_0$.

MAIN RESULT

Recall that three consecutive prime numbers $p_k, p_{k+1}, p_{k+2}$ have the closer property if the differences $d_k$ and $d_{k+1}$ satisfy the condition $d_{k+1} \leq d_k$, an example being 4111, 4127, 4129.

With this refreshed in our minds we now return to our main result.

Theorem 2 (The Three Primes Theorem).

Let the sequence $p_1, p_2,$ ... , $p_n$, ... be the sequence of prime numbers in increasing order. Then there are infinitely many terns of consecutive primes $p_k, p_{k+1}, p_{k+2}$

such that their consecutive differences satisfy the closer condition, that is, $d_{k+1} \leq d_k$

where $d_{k+1} = p_{k+2} - p_{k+1}$ and $d_k = p_{k+1} - p_k$.

Proof. We prove this theorem by Reductio ad absurdum.

Suppose, to get a contradiction, that there are only a finite number of terns satisfying the closer property. Let this last tern be $p_{j-2}, p_{j-1},p_j$

Discarding from the original sequence of prime numbers all primes p $\leq p_j$ (which are finite in number), we obtain the sequence of all the remaining primes p $> p_j$, that is: $p_{j+1}, p_{j+2}, p_{j+3}$, ..., $p_n$, ...

The differences between two consecutive primes of this last sequence must form an increasing sequence. If this last condition is not met, then we have a tern of three consecutive primes $p_r, p_{r+1}, p_{r+2}$ satisfying the closer condition, contradicting the fact that the last tern was $p_{j-2}, p_{j-1},p_j$

Thus, the sequence $p_{j+1}, p_{j+2}, p_{j+3}$, ..., $p_n$, ... satisfies the condition of Theorem 1, that is, their differences form an increasing sequence of natural numbers; then the series of the reciprocals of this sequence is convergent. But the sequence of the reciprocals of the primes, that is: $\sum\limits_{i=1}^{\infty}\frac{1}{p_i}$

is divergent (Euler, 1737). This is the contradiction we were searching for and this completes the proof of our Theorem.

REFERENCES

 I. Niven, H. Zuckerman, H. Montgomery, An introduction to the theory of numbers, 1991 John Wiley \& Sons, Inc.

 G.H. Hardy and E.M. Wright, An introduction to the theory of numbers, Clarendon Press (Oxford), 1979

 L.E. Dickson, History of the theory of numbers, Carnegie Institution of Washington, New York 1950

SPECIAL THANKS TO MY FRIEND AND MENTOR N.H.GUERSENZVAIG FOR REVISING THIS WORK.