Extend Bertrand’s Postulate to Sums of Any Primes

Pham Minh Duc

Department of Physics, VNU University of Science, Hanoi, Vietnam.

DOI: 10.4236/oalib.1110986                

Abstract

According to Bertrand’s postulate, we have P_n+P_n≥P_n+1. Is it true that for all n>1 then P_n-1+P_n≥P_n+1? Then P_n+P_n-i>P_n+j where n≥N, N is a large enough value and i, j are natural numbers?

Keywords

Bertrand’s Postulate, Rosser’s Theorem, L’Hospital Rule, Prime Number

1. Introduction

In 1845, Bertrand conjectured what became known as Bertrand’s postulate: twice any prime strictly exceeds the next prime [1] . Tchebichef presented his proof of Bertrand’s postulate in 1850 and published it in 1852 [2] . It is now sometimes called the Bertrand-Chebyshev theorem. Surprisingly, a stronger statement seems not to be well known, but is elementary to prove: The sum of any two consecutive primes strictly exceeds the next prime, except for the only equality 2 + 3 = 5. After I conjectured and proved this statement independently, a very helpful referee pointed out that Ishikawa published this result in 1934 (with a different proof) [3] . This observation is a special case of a much more general result, Theorem 2, that is also elementary to prove (given the prime number theorem), and perhaps not previously noticed: If $p_n$ denotes the nth prime, $n=1,2,3,\cdots$ with $p_1=2,p_2=3,p_3=5,\cdots$ and if $c_1,c_2,\cdots ,c_j$ are natural numbers (not necessarily distinct), and $d_1,d_2,\cdots ,d_i$ are positive integers (not necessarily distinct), and then there exists a positive integer N such that $p_{n-c_1}+p_{n-c_2}+\cdots +p_{n-c_j}>p_{n+d_1}+p_{n+d_2}+\cdots +p_{n+d_i}$ for ll $n\ge N$ . We also have another result: If $i<n$ and j are nonnegative integers, then there exists a large enough positive integer N such that, for all $n\ge N$ , $p_n+p_{n-i}>p_{n+j}$ . We give some numerical results.

2. Main Result

Theorem 1. If $i<n$ and j are nonnegative integers, then there exists a large enough positive integer N such that, for all $n\ge N$ , $p_n+p_{n-i}>p_{n+j}$ .

Applying Rosser’s theorem for all $n\ge 6$ , we have

$n\left(\ln n+\ln \ln n-1\right)<p_n<n\left(\ln n+\ln \ln n\right)$

$\left(n+j\right)\left[\ln \left(n+j\right)+\ln \ln \left(n+j\right)-1\right]<p_{n+j}<\left(n+j\right)\left[\ln \left(n+j\right)+\ln \ln \left(n+j\right)\right]$

For all $n>i+6$ , we have

$\left(n-i\right)\left[\ln \left(n-i\right)+\ln \ln \left(n-i\right)-1\right]<p_{n-i}<\left(n-i\right)\left[\ln \left(n-i\right)+\ln \ln \left(n-i\right)\right]$

Consider the expression

$A=\frac{n\left(\ln n+\ln \ln n-1\right)+(n-i)\left[\ln \left(n-i\right)+\ln \ln \left(n-i\right)-1\right]}{\left(n+j\right)\left[\ln \left(n+j\right)+\ln \ln \left(n+j\right)\right]}$

We consider the following limit

$B=\underset{n\to +\infty }{\lim }\frac{n\left(\ln n+\ln \ln n-1\right)+\left(n-i\right)\left[\ln \left(n-i\right)+\ln \ln \left(n-i\right)-1\right]}{\left(n+j\right)\left[\ln \left(n+j\right)+\ln \ln \left(n+j\right)\right]}$

Taking the ln of the numerator and denominator and applying L’Hospital Rule gives

$\begin{array}{l}\underset{n\to +\infty }{\lim }n\left(\ln n+\ln \ln n-1\right)\\ =\underset{n\to +\infty }{\lim }\ln n+\ln \ln n-1+n\left(\frac{1}{n}+\frac{\frac{1}{n}}{\ln n}\right)\\ =\underset{n\to +\infty }{\lim }\ln n+\ln \ln n+\frac{1}{\ln n}\end{array}$

$\begin{array}{l}\underset{n\to +\infty }{\lim }n\left[\ln \left(n-i\right)+\ln \ln \left(n-i\right)-1\right]\\ =\underset{n\to +\infty }{\lim }\ln \left(n-i\right)+\ln \ln \left(n-i\right)-1+\left(n-i\right)\left(\frac{1}{n-i}+\frac{\frac{1}{n-i}}{\ln \left(n-i\right)}\right)\\ =\underset{n\to +\infty }{\lim }\ln \left(n-i\right)+\ln \ln \left(n-i\right)+\frac{1}{\ln \left(n-i\right)}\end{array}$

$\begin{array}{l}\underset{n\to +\infty }{\lim }n\left[\ln \left(n+j\right)+\ln \ln \left(n+j\right)\right]\\ =\underset{n\to +\infty }{\lim }\ln \left(n+j\right)+\ln \ln \left(n+j\right)+\left(n+j\right)\left(\frac{1}{n+j}+\frac{\frac{1}{n+j}}{\ln \left(n+j\right)}\right)\\ =\underset{n\to +\infty }{\lim }\ln \left(n+j\right)+\ln \ln \left(n+j\right)+\frac{1}{\ln \left(n+j\right)}+1\end{array}$

Then we see

$B=\underset{n\to +\infty }{\lim }\frac{\ln n+\ln \ln n+\frac{1}{\ln n}+\ln \left(n-i\right)+\ln \ln \left(n-i\right)+\frac{1}{\ln \left(n-i\right)}}{\ln \left(n+j\right)+\ln \ln \left(n+j\right)+1+\frac{1}{\ln \left(n+j\right)}}$

When $n\to +\infty$ then

$B=\underset{n\to +\infty }{\lim }\frac{\ln n+\ln \left(n-i\right)}{\ln \left(n+j\right)}=\underset{n\to +\infty }{\lim }\frac{\ln \left(n^2-in\right)}{\ln \left(n+j\right)}=+\infty$

(Because $n^2-in\gg n+j$ , for $n\to +\infty$ )

Or, for $n\ge N$ , N is a large enough positive integer, then $A>1$ ,

$\frac{n\left(\ln n+\ln \ln n-1\right)+\left(n-i\right)\left[\ln \left(n-i\right)+\ln \ln \left(n-i\right)-1\right]}{\left(n+j\right)\left[\ln \left(n+j\right)+\ln \ln \left(n+j\right)\right]}>1$

It turns out, $p_n+p_{n-i}\ge p_{n+j}$ .

Theorem 2. If $c_1,c_2,\cdots ,c_j$ are j nonnegative integers (not necessarily distinct), and $d_1,d_2,\cdots ,d_i$ are i positive integers (not necessarily distinct), with $1\le i<j$ , then there exists a large enough positive integer N such that, for all $n\ge N$ , $p_{n-c_1}+p_{n-c_2}+\cdots +p_{n-c_j}>p_{n+d_1}+p_{n+d_2}+\cdots +p_{n+d_i}$ .

Applying Rosser’s theorem for all $n\ge 6$ , we have

$\begin{array}{l}\left(n+d_i\right)\left[\ln \left(n+d_i\right)+\ln \ln \left(n+d_i\right)-1\right]<p_{n+d_i}\\ <\left(n+d_i\right)\left[\ln \left(n+d_i\right)+\ln \ln \left(n+d_i\right)\right]\end{array}$

For all $n>c_j+6$ , we have

$\begin{array}{l}\left(n-c_j\right)\left[\ln \left(n-c_j\right)+\ln \ln \left(n-c_j\right)-1\right]<p_{n-c_j}\\ <\left(n-c_j\right)\left[\ln \left(n-c_j\right)+\ln \ln \left(n-c_j\right)\right]\end{array}$

Consider the expression

$C=\frac{{{\displaystyle \sum }}_{g=1}^j\left(n-c_g\right)\left[\ln \left(n-c_g\right)+\ln \ln \left(n-c_g\right)-1\right]}{{{\displaystyle \sum }}_{h=1}^i\left(n+d_h\right)\left[\ln \left(n+d_h\right)+\ln \ln \left(n+d_h\right)\right]}$

We consider the following limit

$D=\underset{n\to +\infty }{\lim }\frac{{{\displaystyle \sum }}_{g=1}^j\left(n-c_g\right)\left[\ln \left(n-c_g\right)+\ln \left(n-c_g\right)-1\right]}{{{\displaystyle \sum }}_{h=1}^i\left(n+d_h\right)\left[\ln \left(n+d_h\right)+\ln \left(n+d_h\right)\right]}$

Taking the ln of the numerator and denominator and applying L’Hospital Rule gives

$\begin{array}{l}\underset{n\to +\infty }{\lim }\underset{g=1}{\overset{j}{{\displaystyle \sum }}}\left(n-c_g\right)\left[\ln \left(n-c_g\right)+\ln \ln \left(n-c_g\right)-1\right]\\ =\underset{n\to +\infty }{\lim }\underset{g=1}{\overset{j}{{\displaystyle \sum }}}\ln \left(n-c_g\right)+\ln \ln \left(n-c_g\right)+\frac{1}{\ln \left(n-c_g\right)}\end{array}$

$\begin{array}{l}\underset{n\to +\infty }{\lim }\underset{h=1}{\overset{i}{{\displaystyle \sum }}}\left(n+d_h\right)\left[\ln \left(n+d_h\right)+\ln \ln \left(n+d_h\right)\right]\\ =\underset{n\to +\infty }{\lim }\underset{h=1}{\overset{i}{{\displaystyle \sum }}}\ln \left(n+d_h\right)+\ln \ln \left(n+d_h\right)+\frac{1}{\ln \left(n+d_h\right)}+1\end{array}$

Then we see

$D=\underset{n\to +\infty }{\lim }\frac{{{\displaystyle \sum }}_{g=1}^j\ln \left(n-c_g\right)+\ln \ln \left(n-c_g\right)+\frac{1}{\ln \left(n-c_g\right)}}{{{\displaystyle \sum }}_{h=1}^i\ln \left(n+d_h\right)+\ln \ln \left(n+d_h\right)+\frac{1}{\ln \left(n+d_h\right)}+1}$

When $n\to +\infty$ then

$D=\underset{n\to +\infty }{\lim }\frac{{{\displaystyle \sum }}_{g=1}^j\ln \left(n-c_g\right)}{{{\displaystyle \sum }}_{h=1}^i\ln \left(n+d_h\right)}=+\infty$

(Because ${\displaystyle {\sum }_{g=1}^j\ln \left(n-c_g\right)}\gg {\displaystyle {\sum }_{h=1}^i\ln \left(n+d_h\right)}$ , for $n\to +\infty$ and $1\le i<j$ )

Or, for $n\ge N$ , N is a large enough positive integer, then $C>1$ ,

$\frac{{{\displaystyle \sum }}_{g=1}^j\left(n-c_g\right)\left[\ln \left(n-c_g\right)+\ln \ln \left(n-c_g\right)-1\right]}{{{\displaystyle \sum }}_{h=1}^i\left(n+d_h\right)\left[\ln \left(n+d_h\right)+\ln \ln \left(n+d_h\right)\right]}>1$

It turns out, $p_{n-c_1}+p_{n-c_2}+\cdots +p_{n-c_j}>p_{n+d_1}+p_{n+d_2}+\cdots +p_{n+d_i}$ .

3. Concluding Remark

In this short note we have provided the prime number inequality via Rosser and Schoenfeld bounds [4] .

Acknowledgements

I thank VNU University of Science for accompanying me.

Conflicts of Interest

The author declares no conflicts of interest.

References

[1]	Bertrand, J. (1845) M’emoire sur le nombre de valeurs que peut prendre une function quand on y permute les lettres qu’elle renferme. Journal de l’Ecole Royale Polytechnique Cahier, 30, 123-140.
[2]	Tchebichef, P. (1852) M’emoire sur les nombres premiers. Journal de Mathématiques pures et appliquées, 17, 366-390.
[3]	Ishikawa, H. (1934) über die Verteilung der Primzahlen. Science Reports of the Tokyo Bunrika Daigaku, Section A, 2, 27-40.
[4]	Rosser, J.B. and Schoenfeld, L. (1962) Approximate Formulas for Some Functions of Prime Numbers. Illinois Journal of Mathematics, 6, 64-94. https://doi.org/10.1215/ijm/1255631807