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

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

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

[3] 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.