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Statement For every $n > 1$ there is always at least one prime $p$ such that $n < p < 2n$.

I am curious to know that if I replace that $2n$ by $2n-\epsilon$, ($\epsilon>0$) then what is the $\inf (\epsilon)$ so that the inequality still holds, meaning there is always a prime between $n$ and $2n-\epsilon$

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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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Bertrand's postulate is

if n > 3 is an integer, then there always exists at least one prime number p with n < p < 2n − 2.

Thus ε < 2 for n > 3. What if n ≤ 3?

  • For n = 3, 3 < 5 < 6 - ε ⇒ ε < 1
  • For n = 2, 2 < 3 < 4 - ε ⇒ ε < 1

Hence we have 0 < ε < 1, if ε is a constant.

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  • $\begingroup$ now the question may be more interesting. I actually wanted to find the least postive $\epsilon$ such that the condition remains true. $\endgroup$ – anonymous Aug 11 '10 at 16:21
  • $\begingroup$ @Chandru1: Any ε between 0 and 1 will do, so the infimum of all possible positive ε is 0. This is not surprising. Did you want the supremum instead? (The supremum is 1 of you want it to hold for n=2 or 3, and infinity if you only want sufficiently large n.) $\endgroup$ – ShreevatsaR Aug 11 '10 at 16:44
  • $\begingroup$ @ShreevatsaR : Hi i got it. $\endgroup$ – anonymous Aug 11 '10 at 16:50
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The answer depends if you want an answer that is true "for all n" or an answer that is true "for all sufficiently large n." For instance, there is not always a prime in an interval of the form (n, 3n/2). Take n=7, for instance. There is always a prime in such an interval "for sufficiently large n," however.

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This was there in the proof of Bertrand's theorem.

if $n>60$, then $\varepsilon=\frac{2n}{3}$.

EDIT: I can't find the exact proof, but almost all proofs include this step:
We prove that $\frac{(2n)!}{((n!)*(n!))}$ has no prime factors at all in the gap interval ($\frac{2n}{3}$ , n)
As a result, if $\varepsilon=\frac{2n}{3}$, then there is no prime number in the interval (n,2n)

Also refer Wiki

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  • $\begingroup$ Hi! Welcome to math.SE. This solution would be more helpful if you cited readers to the proof you are referring to. Might you be able to add that? $\endgroup$ – rschwieb Dec 28 '12 at 14:30
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In 1952, J. Nagura, for example, proved that there is a prime between $n$ and $\frac{6}{5}n$.

https://www.quora.com/Is-Jitsuro-Naguras-1952-proof-that-for-n-geq-25-there-is-always-a-prime-between-n-and-1-+-frac-1-5-n-generally-accepted

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