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Define $q:\mathbb{N}\rightarrow\mathbb{Q}^+$ with, $q(n)=min\{\alpha\in \mathbb{Q}^+\mid \exists \;p\in \mathbb{N},\; p \text{ prime and } p \in (n,\alpha n]\}$.

Is the $\liminf\limits_{n\rightarrow\infty} q(n)=1$?

I was thinking that for $p$ prime, $q(p-1)=1+\frac{1}{p-1}$ as $(1+\frac{1}{p-1})(p-1)=p$ is prime. Then this tends to $1$ for large primes $p$.

Also does Bertrand's postulate imply that $\liminf\limits_{n\rightarrow\infty} q(n)\leq 2$?

I'm just curious really, I was reading over bertrand's postulate and thought of this.

snulty
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  • Your example of $p-1$ shows that $\liminf\limits_{n\to\infty} q(n) = 1$. Bertrand's postulate/Chebyshev's theorem shows $\limsup\limits_{n\to\infty} q(n) \leqslant 2$. The interesting question would be whether $\limsup\limits_{n\to\infty} q(n) = 1$. I think we have results implying that, but I may misremember, not my department of expertise. – Daniel Fischer May 18 '14 at 20:31
  • @DanielFischer oh yeah, I see what you mean, lim sup makes sense. I got confused. Thanks – snulty May 18 '14 at 20:36

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yes the limit is 1. can i ask you: why you need this ?

adddddddddddd
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  • I don't need it. I was just curious if I understood lim inf and lim sup properly. Apparently I don't :) – snulty May 18 '14 at 20:44