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

dourtuntellorvl 2022-06-29 Answered
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?
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Answers (1)

Answered 2022-06-30 Author has 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.
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