Suppose that we choose some real number $\varepsilon >0$.
Can we always find $n_0(\varepsilon) \in \mathbb N$ such that for every $n> n_0(\varepsilon)$ there is a prime number $p$ such that we have $n<p<n(1+\varepsilon)$?
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Farewell
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Yes, and this is well-known - see the "Better Results" section of https://en.wikipedia.org/wiki/Bertrand's_postulate , for instance. – Steven Stadnicki May 20 '16 at 22:46
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@StevenStadnicki Nice. You could post that as an answer? – Farewell May 20 '16 at 22:49
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Oh, we have an answer in the same spirit. – Farewell May 20 '16 at 22:51
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dupe of http://math.stackexchange.com/questions/1090770/a-question-on-prime-density/1090816#1090816, sort of. – daniel May 21 '16 at 08:22
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yes, see: https://en.wikipedia.org/wiki/Bertrand%27s_postulate --> "Better results"
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