Finding liminf of p_n/n where p_n = nth prime

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Homework Help Overview

The discussion revolves around finding the limit inferior of the sequence defined by p_n/n, where p_n represents the nth prime number. Participants explore the asymptotic behavior of prime numbers and their relationship to n, particularly through the approximation p_n ~ n log n.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of the asymptotic relation p_n ~ n log n and question the validity of substituting this into the limit inferior expression. There is also consideration of whether the sequence n/p_n is monotonic for large n and how this affects the limit.

Discussion Status

Some participants have provided insights into the implications of the asymptotic relationship and have justified certain substitutions. However, questions remain regarding the monotonicity of the sequence n/p_n and its implications for the limit inferior.

Contextual Notes

There is an ongoing discussion about the nature of the sequence and whether certain conditions hold true for large n, particularly in relation to the behavior of the primes. Participants are also considering the implications of their findings on the original question.

Dragonfall
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Homework Statement



Find the lim inf of p_n/n where p_n is the nth prime.

Homework Equations



Well p_n ~ n logn, but I'm not sure if a simple substitution would work. This question may be incredibly trivial or open, and I can't figure out which.

I'm also wondering if the sequence above is monotone decreasing for sufficiently large n (this is not true for small n).
 
Last edited:
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You have to think what [tex]P_n[/tex]~ [tex]n \log n[/tex] actually means!

It means the quotient of the 2 functions as n approaches infinity is 1, note this does not mean the difference of the functions as n approaches infinity is 0.

eg n+2~n, but the difference of the functions will always be 2.

Anywho, so we know [tex]\lim_{n\rightarrow {\infty}} \frac{P_n}{n\log n} = 1[/tex].

Since we want [tex]\lim_{n\rightarrow {\infty}} \frac{P_n}{n}[/tex], we can make the simple substitution and get log n.
 
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Ok I wrote that backwards. I want n/p_n, not p_n/n. The question is whether lim n/(nlog n)=0 implies that lim inf n/p_n=0. More importantly is whether n/p_n is monotone for large n.
 
Well as per my previous post, the substitution is justified. So Your first question is true, it implies it. The 2nd part, I would not know.
 
An equivalent question is whether [tex]np_{n+1}-(n+1)p_n<0[/tex] for only finitely many n.
 

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