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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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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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  • $\begingroup$ 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. $\endgroup$
    – Ilya Nikokoshev
    Commented Oct 26, 2009 at 23:55
  • $\begingroup$ ~ 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. $\endgroup$
    – David E Speyer
    Commented Oct 26, 2009 at 23:59
  • $\begingroup$ Mm, yes it is so. $\endgroup$
    – Ilya Nikokoshev
    Commented Oct 27, 2009 at 0:04
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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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