Recall that Bertrand's postulate states that for n &#x2265;<!-- ≥ --> 2 there always exi

dourtuntellorvl

dourtuntellorvl

Answered question

2022-06-29

Recall that Bertrand's postulate states that for n 2 there always exists a prime between n and 2 n. Bertrand's postulate was proved by Chebyshev. Recall also that the harmonic series
1 + 1 2 + 1 3 + 1 4 +
and the sum of the reciprocals of the primes
1 2 + 1 3 + 1 5 + 1 7 +
are divergent, while the sum
n = 0 1 n p
is convergent for all p > 1. This would lead one to conjecture something like:

For all ϵ > 0, there exists an N such that if n > N, then there exists a prime between n and ( 1 + ϵ ) n.

Question: Is this conjecture true? If it is true, is there an expression for N as a function of ϵ?

Answer & Explanation

odmeravan5c

odmeravan5c

Beginner2022-06-30Added 20 answers

That result follows from the Prime Number Theorem. You can make it effective with a result of Dusart: for n 396738 , there is always a prime between n and n + n / ( 25 log 2 n ). So in particular this holds for
N max ( exp ( 1 25 ε ) ,   396738 ) .
On the Riemann hypothesis (using the result of Schoenfeld) there is a prime between x log 2 x x 4 π and x for x 599 and this should give a better bound.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Elementary geometry

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?