## 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)