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Feb 9, 2018 at 17:54 history edited GH from MO CC BY-SA 3.0
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Feb 9, 2018 at 17:17 comment added user1582006 @NoamD.Elkies I know the simplest proofs of Bertrand's Postulate from Erdos and Ramanujan, but actually my proof is shorter and simpler.
Feb 9, 2018 at 17:14 comment added Noam D. Elkies Meanwhile, yes, "Bertrand's Postulate" was already proved 100+ years ago by Chebyshev, with a much more elementary proof than any known for the Prime Number Theorem (the key tool is the formula for the prime factorization of $x!$).
Feb 9, 2018 at 17:01 history edited GH from MO CC BY-SA 3.0
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Feb 9, 2018 at 16:38 history edited GH from MO CC BY-SA 3.0
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Feb 9, 2018 at 16:31 comment added user1582006 @DustanLevenstein Yes. I've made a change. Now I correct this one accordingly.
Feb 9, 2018 at 16:31 comment added Dustan Levenstein @user1582006 Your claim is "for every $n>p$ something happens". That includes the possibility that $n$ not be a prime.
Feb 9, 2018 at 16:28 comment added user1582006 @GHfromMO Who said that "n" could not be a prime???
Feb 9, 2018 at 16:27 comment added user1582006 @GHfromMO Please reread my question. There was an error.
Feb 9, 2018 at 16:24 comment added user1582006 @GHfromMO You are right. I've made an error by transcribing. Actually it's less than or equal to. See my "edit" on the question.
Feb 9, 2018 at 16:18 comment added user1582006 @GHfromMO Please note that for ε = 1, I've a very simple and short proof of Bertrand's Postulate. Do you think that one will be a result worth to publish?
Feb 9, 2018 at 16:14 comment added GH from MO @user1582006: $\varepsilon$ is arbitrary, and $n$ has to be sufficiently large in terms of it. How large $n$ has to be is a different question: there are versions of the prime number theorem with good explicit error terms, which can be used for that purpose. See, for example, iml.univ-mrs.fr/~ramare/TME-EMT/Articles/Art01.html
Feb 9, 2018 at 16:06 comment added user1582006 @GHfromMo But how do you compute ϵ? With my inequality you can actually find that!
Feb 9, 2018 at 16:04 history edited GH from MO CC BY-SA 3.0
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Feb 9, 2018 at 16:02 comment added GH from MO @user1582006: Here is the proof: $\pi(n)=(1+o(1))n/\log n$ and $\pi((1+\varepsilon)n)=(1+\epsilon+o(1))n/\log n$. Now subtract the first expression from the second one.
Feb 9, 2018 at 16:01 comment added user1582006 @Noam_D._Elkies Ok. But where can I find a proof of that?
Feb 9, 2018 at 16:01 comment added GH from MO @user1582006: It follows immediately from the prime number theorem. For any fixed $\varepsilon>0$, the number of primes between $n$ and $(1+\varepsilon)n$ is asymptotically $\varepsilon n/\log n$. For similar results in much shorter intervals, see Section 10.5 in Iwaniec-Kowalski: Analytic number theory.
Feb 9, 2018 at 16:00 comment added Noam D. Elkies It's a well-known easy consequence of the Prime Number Theorem.
Feb 9, 2018 at 15:57 comment added user1582006 @GH_from_MO "It is also-well known"? Where can I get a proof of that?
Feb 9, 2018 at 15:54 history answered GH from MO CC BY-SA 3.0