## A new result on the frequency with which Erdos and Turan inequalities occur

May 10th, 2014 | Posted by in - (Comments Off on A new result on the frequency with which Erdos and Turan inequalities occur)

Author : MARTIN MAULHARDT

ABSTRACT.

In our previous two papers we have proved by a different and simpler method than the one used by Erdos and Turan that the inequality  $p_{k+1} - p_k \leq p_k - p_{k-1}$ is satisfied infinitely many times in the set of prime numbers. We have also extend this property to any large set. That is, these inequalities are a consequence of the divergence of the series of the reciprocals of any large set. It is not a consequence of the fact that any number is a unique product of primes. In our second paper we continued the work of Erdos and Turan proving that the more general inequalities  $p_{k+r} - p_k \leq p_k - p_{k-r}$ of which the case $r=1$ is just an example are satisfied infinitely many times for any large set (in particular the set of prime numbers) for every $r \in N$. Unfortunately Erdos and Turan did not consider the problem of the frequency with which these inequalities occur. We have found an answer and it is on this answer that this paper is all about.

Note : The reader might want to check our first two works on this same webpage if he does not feel familiar with some theorems and/or definitions.

1. Introduction and Basic Definitions

Definition 1. A triple of natural numbers $(a,b,c)$ is said to have the closer property if $c-b \leq b-a$.

For example, the triple $(7,11,13)$ has the closer property. So does the triple $(23,29,31)$

The convergence criteria and the three closer primes theorem of our first work of this webpage prove that :

Theorem 2. The set $P$ of prime numbers has infinitely many triples of consecutive prime numbers with the closer property.

The fact that these triples are infinite is a good result but it demands more precision. For example, if we put in correspondence all triples of consecutive prime numbers with the natural numbers i.e. we labeled :

$T_1 = (2,3,5)$

$T_2 = (3,5,7)$

$T_3 = (5,7,11)$

..................................

$T_n = (p_n,p_{n+1},p_{n+2})$

some of these will have the closer property and some will not. It might happen that, starting at some point on, the triples with the closer property are very unusual. For example they might be only the followings : $T_{100} , T_{10,000} , T_{1,000,000} , T_{100,000,000} ........... T_{100^{n}} .........$ .

In this case the indexes of the triples with the closer property are in geometric rate.

Or it might happen that they are the followings : $T_{100} , T_{121} , T_{144} , T_{169} ........... T_{n^2}...........$.

In both cases they are infinite and the reader of this paper realizes at once that, in some sense, in the second case we have more triples with the closer property than in the first.

We now extend the definition of the closer property for some sequences of natural numbers.

Definition 3. Let $a_n$ be an infinite strictly increasing sequence of natural numbers  say $A = (a_1, a_2, \dots a_n \dots )$. We say that A has the closer property if there are infinite triples of consecutive elements of A with the closer property.

For example, the sequence, $a_n=(1,2,4,5,7,8, \dots 3n+1,3n+2, \dots)$ has the closer property. So does the sequence of prime numbers and any large set considered as a sequence.

2. Main result of this paper

We are now ready to state and prove the main result of this paper.

We will prove that in the sequence of prime numbers $P = (2,3,5, \dots ,p_n, \dots )$ the set of indexes $I$ with triples satisfying the closer property ALSO has the closer property. That is, there is no $n_0 \in N$ from which starting at that $n_0$ the indexes of the triples with the closer property separate more and more.

Let us consider the following sequence of increasing natural numbers :

$a_n = (1, 3; 4, 6, 9; 10, 12, 15, 19; 20, 22, 25, 29, 34; \dots)$

This sequence if formed by the following law :

The differences between consecutive terms are

$d_n = (2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6,\dots)$

We now prove a Lemma related to this sequence $a_n$

Lemma 4. The series of the reciprocals of $a_n$ is convergent, i.e :

$\sum\limits_{n=1}^{\infty}\frac{1}{a_n} < \infty$

Proof. We write the series in the explicit form

$1 + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{19} +\dots$.

Let us now associate its terms in the following way :

$(1 + \frac{1}{3}) + (\frac{1}{4} + \frac{1}{6} + \frac{1}{9}) + (\frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{19}) +\dots$.

We have 2 terms in the first parenthesis, 3 terms in the second parenthesis, 4 terms in the third parenthesis, etc. Using the comparison criteria we found that this series  sums less than the following series :

$(1 + 1) + (\frac{1}{4} + \frac{1}{4} + \frac{1}{4}) + (\frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10}) +\dots =$.

$2 + \frac{3}{4} + \frac{4}{10} + \frac{5}{20} +\dots +\frac{n+1}{\frac{n^3+3n^2+2n}{6}} +\dots =$ $\sum\limits_{n=1}^{\infty}\frac{6(n+1)}{n^3+3n^2+2n} < \infty$

and this concludes the proof of the lemma.

We are now ready to prove the following important theorem :

Theorem 5. Let $A=(a_1, a_2,\dots ,a_n,\dots )$ be such that $\sum\limits_{n=1}^{\infty}\frac{1}{a_n} = \infty$. Let $T_n= (a_n,a_{n+1},a_{n+2})$. Let the indexes of the infinitely many triples $T_n$ with the closer property be $I = (i_o,i_1,i_2, \dots ,i_n, \dots )$. Then $I$ also has the closer property, i.e.  there exists infinitely many triples $(i_n,i_{n+1},i_{n+2})$ such that $i_{n+2} - i_{n+1} \leq i_{n+1} - i_n$.

Proof. Suppose that starting at some $k \in N$ all indexes of triples with the closer property satisfy the inequality $i_{n+2} - i_{n+1} > i_{n+1} - i_n$. Let the triple $T_k$ be $(a_k,a_{k+1},a_{k+2})$. Consider the sequence of the $a_n$ starting at $k$. Let this sequence be : $a_k,a_{k+1},a_{k+2}, \dots$. (We assume $k>3$). Now $a_k \geq 1,a_{k+1}\geq 3,a_{k+2}\geq 4, a_{k+3}\geq 6,a_{k+4}\geq 9,a_{k+5}\geq 10, \dots$.

The reciprocals of this $a_k$ form a convergent series by comparing this series with the one of lemma 4. But the series of the reciprocals of the $a_n$ starting at $n=1$ is divergent. Absurd! Thus the set of the indexes of triples with the closer property ALSO has the closer property.

Corollary 6. Let $P$ be the set of prime numbers. Let the infinite triples of primes with the closer property be $T_{i_0},T_{i_1},T{i_2},\dots T_{i_n},\dots$. Then there exists infinitely many $i_n$ such that $i_{n+2}-i_{n+1}\leq i_{n+1}-i_n$.

## A generalization of an inequality of Erdos and Turan

September 8th, 2013 | Posted by in - (Comments Off on A generalization of an inequality of Erdos and Turan)

Author: MARTIN MAULHARDT

ABSTRACT.

In 1948 Paul Erdos and Pal Turan proved in their paper : "On some new questions on the distribution of prime numbers" that the inequality $p_{k+2} - p_{k+1} < p_{k+1} - p_k$ is satisfied infinitely many times in the set of prime numbers. Their inequality can be interpreted geometrically by saying that the distance to the prime to the right of $p_{k+1}$ is smaller than the distance to the prime to the left of $p_{k+1}$ infinitely many times.  They did a great job. Their analysis however is incomplete. They could have asked if the second prime to the right of a prime is nearer than the second prime to the left. Or the third one, or the n-th one. We will prove, using only divergent series, that this is the case, that is : the inequality $p_{k+4} - p_{k+2} \leq p_{k+2} - p_k$ is satisfied infinitely many times in the set of prime numbers. For example the two primes that follow 41 to the right are 43 and 47 and the two to the left are 31 and 37. The difference between 41 and 47 is six, smaller than the difference between 41 and 31 which is ten. This fact is true with the third prime, the fourth prime, and in general with the n-th prime infinitely many times.

INTRODUCTORY THEOREMS

In order to make this article self-contained we repeat here the exact formulation of the convergence criteria, the three closer primes theorem and their corresponding proofs.

Theorem 1. (Convergence Criteria) 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. ||||

For example, consider the Fibonacci sequence. 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, 3, 5, 8, 13 ....

These 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 differences form a strictly increasing sequence by the weaker condition that they form a strictly increasing sequence starting at some point $n_0$.

Theorem 2. (The Three Closer 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. ||||

Note : The Three Closer Primes Theorem is one of the inequalities in the work of Erdos and Turan, but we observe that our proof is completely different and simpler than theirs. Our proof can be applied to other sets and not just to the whole set of primes. This is the key idea of our work.

LARGE SETS AND THEIR CONDITIONS

We now define the concept of large set.

Definition. A set {$a_n$} of positive integers is said to be a large set if the series of its reciprocals is divergent, i.e., $\sum\limits_{n=1}^{\infty}\frac{1}{a_n} = \infty$.

We have several examples of large sets :

1) The whole natural numbers N is a large set.

2) Any arithmetic progression is a large set.

3) The set of primes is a large set.

4) The set of square numbers is NOT a large set.

Definition. Let $p_1$, $p_2$, $p_3$, ... , $p_n$, ... denote the sequence of prime numbers. Then the subsequence $p_1$, $p_3$, $p_5$, ... , $p_{2k+1}$, ... is the odd indexed primes set and the subsequence $p_2$, $p_4$, $p_6$, ... , $p_{2k}$ , ... is the even indexed primes set.

Lemma 1. If $p_n$ denotes the sequence of primes ordered in increasing magnitude then the odd indexed primes form a large set and so does the even indexed primes.

Proof

If the sequence of even indexed primes is not a large set, i.e, if the series $\sum\limits_{k=1}^{\infty}\frac{1}{p_{2k}}$ is convergent then the series $\sum\limits_{k=1}^{\infty}\frac{1}{p_{2k+1}}$ is also convergent by the comparison criteria. Then the series $\sum\limits_{k=1}^{\infty}\frac{1}{p_k}$ is convergent. But this series is divergent (Euler, 1737).

The same proof works if we start with the odd indexed primes. ||||

Definition. A tern of positive integers $(a, b, c)$ is said to satisfy the closer property if $c-b \leq b-a.$

For example $(7, 11, 13)$ satisfies the closer property.

We now restate theorem 1 in its counter reciprocal form which shows a necessary condition for a set to be large and name it theorem 3.

Theorem 3. Let $a_n$ be a sequence of positive integers ordered in increasing magnitude. Let the consecutive differences be

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

If the series $\sum\limits_{n=1}^{\infty}\frac{1}{a_n}$ is divergent then their differences satisfy infinitely many times the inequalities

$d_{n+1} \leq d_n$

i.e., the closer property

$a_{n+2} - a_{n+1} \leq a_{n+1} - a_n$

is satisfied infinitely many times.

Proof.

This is just the counter reciprocal of theorem 1. ||||

We now combine theorem 3 with lemma 1 to obtain the main result mentioned in the abstract :

Theorem 4. Let $p_2$, $p_4$, $p_6$, ..... , $p_{2k}$ , ..... be the sequence of even indexed primes.

Then the inequality

$p_{2k+4} - p_{2k+2} \leq p_{2k+2} - p_{2k}$

is satisfied infinitely many times.

Proof.

By lemma 1 the set of even indexed primes is a large set. Then the series $\sum\limits_{n=1}^{\infty}\frac{1}{p_{2k}}$ is divergent.

Then by theorem 3 the inequality

$p_{2k+4} - p_{2k+2} \leq p_{2k+2}$ - $p_{2k}$

is satisfied infinitely many times. ||||

Theorem 4 points out some interesting facts. Consider the distance between $p_{2k}$ and $p_{2k+2}$, then in that same distance from $p_{2k+2}$ on, we will eventually find the next two primes $p_{2k+3}$ and $p_{2k+4}$ and it is easy to see how this can be generalized to the next $j$ primes: in the distance between $p_{k}$ and $p_{k+j}$, if we continue that same distance from $p_{k+j}$ on, we will eventually find the next $j$ primes. We have just to consider instead of odd and even indexed primes, primes of the classes $p_{3k}$, $p_{3k+1}$ and $p_{3k+2}$. And so on for any finite j, for example 1,000,000 primes.

Also note that these inequalities are true infinitely many times for any large set. We are not using the fact that the $p_n$ are primes in their proofs.

Note to the last that this theorem asserts that the consecutive differences between even indexed primes satisfy an analogous inequality to the one find by Erdos and Turan for consecutive primes. They found that infinitely many times $p_{n+2} - p_{n+1} \leq p_{n+1} - p_n$ and we found that this inequality is satisfied infinitely many times taking step of two primes, and in general taking steps between k primes.

In order to formally state this facts we need to define two more integers subsets.

Definition. Let $a < b$ be positive integers. Then the set $(( a , b ))$ is the set formed by all the positive integers n satisfying a < n < b. And $((a , b ]]$ is the set formed by all positive integers n satisfying $a < n \leq b$.

For example (( $3 , 7$ )) = {$4, 5, 6$}. There are b-a-1 integers in this set. And ((3 , 7 ]] = {$4, 5, 6, 7$}. There are b-a integers in this set.

With the definitions of these sets we formally write down the observation following theorem 4 in the next corollary.

Corollary. Let $p_{2k}, p_{2k+2}, p_{2k+4}$ be three consecutive even indexed primes satisfying the inequality of Theorem 4. Let

$l=p_{2k+2}$ - $p_{2k}$

then $p_{2k+3}$ and $p_{2k+4}$ both belong to the set

$((p_{2k+2},p_{2k+2}+l]]$

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

[4] P. Erdos and P. Turan, On some new questions on the distribution on prime numbers, Bull. Amer. Math. Soc. Vol 54, number 4 (1948)

## A comment from Gergely Harcos

July 12th, 2013 | Posted by in Opinion - (0 Comments)

On July the 10 th I have posted on Terry Tao 's webpage page my result on the three primes theorem. A few hours later Gergely Harcos from the Alfred Renyi institute pointed out that my result already existed. It was proved by Paul Erdos and Paul Turan in 1948. Here is the link of this remarkable paper, particularly inequalities (7).

http://www.ams.org/journals/bull/1948-54-04/S0002-9904-1948-09010-3/S0002-9904-1948-09010-3.pdf

When I published this theorem as mine in my website I was not aware that these great mathematicians had preceded me. But you may judge my good faith based on the fact that my proof is far simpler and is based on an entirely different construction.

As always I 'm not trying to compare myself with these great mathematicians by saying that my proof is simpler and based on an entirely different construction.

Thanks once again to Gergely Harcos.

## The Three Primes Theorem

July 11th, 2013 | Posted by in - (Comments Off on 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.