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Is it known that for every epsilon there is N_0 such that all intervals of the form [N, (1+\epsilon)*N], where N > N_0, contain prime numbers?

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up vote 8 down vote accepted

This follows from the Prime Number Theorem. Let π(n) be the number of primes less than n. Then π(n) ~ n/log(n); it follows π((1+ε)n)-π(n) -> ∞ as n -> ∞.

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How string is the statement that π(n) ~ n/log(n)? One certainly imagines it to be strong enough to say something specific about pi(n)-pi(n-1), but this is certainly not the typical Prime Number Theorem people study. – Ilya Nikokoshev Oct 26 '09 at 23:55
~ has a precise definition: f(n) ~ g(n) means that lim_{n -> infty} f(n)/g(n)=1. That's strong enough to derive the desired conclusion. – David Speyer Oct 26 '09 at 23:59
Mm, yes it is so. – Ilya Nikokoshev Oct 27 '09 at 0:04

If one wants an explicit bound on N0, apparently this can be gleaned from a Ph.D. thesis by Pierre Dusart (in French) which contains the result that for all x > 3275 there is a prime between x and x(1 + 1/(2 ln2 x)). So we can take N0 to be max(3275,exp((2 epsilon)−1/2)).

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(Google Translate translates the thesis but the equations get messed up.) The statement can be found on and the same is derived near the end of the…. – Unknown Feb 14 '11 at 0:00

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