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 , ⋯ with p 1 = 2 , p 2 = 3 , p 3 = 5 , ⋯ and if c 1 , c 2 , ⋯ , c j are natural numbers (not necessarily distinct), and d 1 , d 2 , ⋯ , 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 + ⋯ + p n − c j > p n + d 1 + p n + d 2 + ⋯ + p n + d i for ll n ≥ 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 ≥ N , p n + p n − i > p n + j . We give some numerical results.

Theorem 1. If i < n and j are nonnegative integers, then there exists a large enough positive integer N such that, for all n ≥ N , p n + p n − i > p n + j .

Applying Rosser’s theorem for all n ≥ 6 , we have

n ( ln n + ln ln n − 1 ) < p n < n ( ln n + ln ln n )

( n + j ) [ ln ( n + j ) + ln ln ( n + j ) − 1 ] < p n + j < ( n + j ) [ ln ( n + j ) + ln ln ( n + j ) ]

For all n > i + 6 , we have

( n − i ) [ ln ( n − i ) + ln ln ( n − i ) − 1 ] < p n − i < ( n − i ) [ ln ( n − i ) + ln ln ( n − i ) ]

Consider the expression

A = n ( ln n + ln ln n − 1 ) + ( n − i ) [ ln ( n − i ) + ln ln ( n − i ) − 1 ] ( n + j ) [ ln ( n + j ) + ln ln ( n + j ) ]

We consider the following limit

B = lim n → + ∞ n ( ln n + ln ln n − 1 ) + ( n − i ) [ ln ( n − i ) + ln ln ( n − i ) − 1 ] ( n + j ) [ ln ( n + j ) + ln ln ( n + j ) ]

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

lim n → + ∞ n ( ln n + ln ln n − 1 ) = lim n → + ∞ ln n + ln ln n − 1 + n ( 1 n + 1 n ln n ) = lim n → + ∞ ln n + ln ln n + 1 ln n

lim n → + ∞ n [ ln ( n − i ) + ln ln ( n − i ) − 1 ] = lim n → + ∞ ln ( n − i ) + ln ln ( n − i ) − 1 + ( n − i ) ( 1 n − i + 1 n − i ln ( n − i ) ) = lim n → + ∞ ln ( n − i ) + ln ln ( n − i ) + 1 ln ( n − i )

lim n → + ∞ n [ ln ( n + j ) + ln ln ( n + j ) ] = lim n → + ∞ ln ( n + j ) + ln ln ( n + j ) + ( n + j ) ( 1 n + j + 1 n + j ln ( n + j ) ) = lim n → + ∞ ln ( n + j ) + ln ln ( n + j ) + 1 ln ( n + j ) + 1

Then we see

B = lim n → + ∞ ln n + ln ln n + 1 ln n + ln ( n − i ) + ln ln ( n − i ) + 1 ln ( n − i ) ln ( n + j ) + ln ln ( n + j ) + 1 + 1 ln ( n + j )

When n → + ∞ then

B = lim n → + ∞ ln n + ln ( n − i ) ln ( n + j ) = lim n → + ∞ ln ( n 2 − i n ) ln ( n + j ) = + ∞

(Because n 2 − i n ≫ n + j , for n → + ∞ )

Or, for n ≥ N , N is a large enough positive integer, then A > 1 ,

n ( ln n + ln ln n − 1 ) + ( n − i ) [ ln ( n − i ) + ln ln ( n − i ) − 1 ] ( n + j ) [ ln ( n + j ) + ln ln ( n + j ) ] > 1

It turns out, p n + p n − i ≥ p n + j .

Theorem 2. If c 1 , c 2 , ⋯ , c j are j nonnegative integers (not necessarily distinct), and d 1 , d 2 , ⋯ , d i are i positive integers (not necessarily distinct), with 1 ≤ i < j , then there exists a large enough positive integer N such that, for all n ≥ N , p n − c 1 + p n − c 2 + ⋯ + p n − c j > p n + d 1 + p n + d 2 + ⋯ + p n + d i .

Applying Rosser’s theorem for all n ≥ 6 , we have

( n + d i ) [ ln ( n + d i ) + ln ln ( n + d i ) − 1 ] < p n + d i < ( n + d i ) [ ln ( n + d i ) + ln ln ( n + d i ) ]

For all n > c j + 6 , we have

( n − c j ) [ ln ( n − c j ) + ln ln ( n − c j ) − 1 ] < p n − c j < ( n − c j ) [ ln ( n − c j ) + ln ln ( n − c j ) ]

Consider the expression

C = ∑ g = 1 j ( n − c g ) [ ln ( n − c g ) + ln ln ( n − c g ) − 1 ] ∑ h = 1 i ( n + d h ) [ ln ( n + d h ) + ln ln ( n + d h ) ]

We consider the following limit

D = lim n → + ∞ ∑ g = 1 j ( n − c g ) [ ln ( n − c g ) + ln ( n − c g ) − 1 ] ∑ h = 1 i ( n + d h ) [ ln ( n + d h ) + ln ( n + d h ) ]

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

lim n → + ∞ ∑ g = 1 j ( n − c g ) [ ln ( n − c g ) + ln ln ( n − c g ) − 1 ] = lim n → + ∞ ∑ g = 1 j ln ( n − c g ) + ln ln ( n − c g ) + 1 ln ( n − c g )

lim n → + ∞ ∑ h = 1 i ( n + d h ) [ ln ( n + d h ) + ln ln ( n + d h ) ] = lim n → + ∞ ∑ h = 1 i ln ( n + d h ) + ln ln ( n + d h ) + 1 ln ( n + d h ) + 1

Then we see

D = lim n → + ∞ ∑ g = 1 j ln ( n − c g ) + ln ln ( n − c g ) + 1 ln ( n − c g ) ∑ h = 1 i ln ( n + d h ) + ln ln ( n + d h ) + 1 ln ( n + d h ) + 1

When n → + ∞ then

D = lim n → + ∞ ∑ g = 1 j ln ( n − c g ) ∑ h = 1 i ln ( n + d h ) = + ∞

(Because ∑ g = 1 j ln ( n − c g ) ≫ ∑ h = 1 i ln ( n + d h ) , for n → + ∞ and 1 ≤ i < j )

Or, for n ≥ N , N is a large enough positive integer, then C > 1 ,

∑ g = 1 j ( n − c g ) [ ln ( n − c g ) + ln ln ( n − c g ) − 1 ] ∑ h = 1 i ( n + d h ) [ ln ( n + d h ) + ln ln ( n + d h ) ] > 1

It turns out, p n − c 1 + p n − c 2 + ⋯ + p n − c j > p n + d 1 + p n + d 2 + ⋯ + p n + d i .

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

I thank VNU University of Science for accompanying me.

The author declares no conflicts of interest.

Duc, P.M. (2023) Extend Bertrand’s Postulate to Sums of Any Primes. Open Access Library Journal, 10: e10986. https://doi.org/10.4236/oalib.1110986