Statement For every n > 1 there is always at least one prime p such that n

lilmoore11p8 2022-07-02 Answered
Statement For every n > 1 there is always at least one prime p such that n < p < 2 n.
I am curious to know that if I replace that 2 n by 2 n ϵ, ( ϵ > 0) then what is the inf ( ϵ ) so that the inequality still holds, meaning there is always a prime between n and 2 n ϵ
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Answers (2)

Answered 2022-07-03 Author has 20 answers
Three related points are worthy of mention, showing that epsilon can be close to n.

There is a result of Finsler that approximates how many primes lie between n and 2n, which is of order o(n/log(n)) as is to be expected by the Prime Number Theorem.

Literature on prime gaps will tell you the exponent delta such that there is (for sufficiently large n) at least one prime in the interval (n , n + n^delta). I think delta is less than 11/20.

Observed data suggests that n^delta can be replaced by something much smaller: for n between something like 3 and 10^14 , some function like 2(log(n))^2 works
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Ayaan Barr
Answered 2022-07-04 Author has 6 answers
Bertrand's postulate is
if n > 3 is an integer, then there always exists at least one prime number p with n < p < 2 n 2.
Thus ξ < 2 for n > 3. What if n 3?

For n = 3 , 3 < 5 < 6 ξ ξ < 1
For n = 2 , 2 < 3 < 4 ξ ξ < 1

Hence we have 0 < ξ < 1, if ε is a constant.
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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
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are divergent, while the sum
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