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*not*monotone for any n, but I can't give a proof of it without assuming something at least as strong as the twin prime conjecture. I was wondering if anyone has some advice to prove this using known methods?

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- #2

mathman

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Statement is confusing - what is variable (not n from what you said)?is not monotone for any n

- #3

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The variable is n. I mean no tail of the sequence is monotone.

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Any ideas at all? I'm drawing dead here.

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CRGreathouse

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I'm still not sure what you mean, exactly.Any ideas at all? I'm drawing dead here.

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The tail of a sequence [tex]a_n[/tex] is the subsequence [tex](a_n)_{n>N}[/tex] for some N. A sequence may not be monotone for the first N numbers, but the tail of the sequence might be monotone. If [tex]\frac{n}{p_n}[/tex] is eventually monotone, then the alternating series test shows that [tex]\sum_n(-1)^n\frac{n}{p_n}[/tex] converges. I believe that the series converges, but I don't think the sequence is eventually monotone, mainly because I think the twin prime conjecture is true.

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- #7

Gib Z

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2, 3, 5, 3, 5, 6, 7, 6, 4, 2, 1/11, 1/12, 1/13, 1/14, 1/15, 1/16, 1/17, 1/18, ...

This sequence is monotone decreasing from the 11th entry onwards.

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Have you tryed to link it with http://mathworld.wolfram.com/PrimeNumberTheorem.html" [Broken]?

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All I can conclude from PNT is that [tex]\lim\frac{n}{p_n}=0[/tex].

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I think it's darn difficult to prove that!

Need to know of distribution of primes for every finite segment [n,n+k] of natural numbers.Looks intractible at first glance.Sorry.

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No, it doesn't.If [tex]\frac{n}{p_n}[/tex] is eventually monotone, then the alternating series test shows that [tex]\sum_n(-1)^n\frac{n}{p_n}[/tex] converges.

- #13

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Why not? [itex]\lim_{n\to \infty}\frac{n}{p_n}=0[/itex] by the PNT, so wouldn't eventual monotonicity be sufficient for convergence?No, it doesn't.

Anyway, let's just see what we have.

No tail is monotone if for all natural numbers N, there is an n>N such that

[tex]\frac{n}{p_n} < \frac{n+1}{p_{n+1}}[/tex].

This is equivalent to

[tex]np_{n+1}<np_n+p_n[/tex]

[tex]p_{n+1}-p_n<\frac{p_n}{n}[/tex]

You don't need the twin primes conjecture here. It suffices to have this lemma:

Even this is stronger than you need, but much, much weaker than the twin primes conjecture.

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Man do I feel stupid.

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Thanks, and you're right, something like "there exists infinitely many n such that [tex]p_{n+1}-p_n<\log n[/tex]" is approximately what I need.This is equivalent to

[tex]np_{n+1}<np_n+p_n[/tex]

[tex]p_{n+1}-p_n<\frac{p_n}{n}[/tex]

You don't need the twin primes conjecture here. It suffices to have this lemma:

There exists a natural number k such that there are infinitely many consecutive primes whose difference is less than k.

- #16

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Anyway, I did some looking, and I found this paper, which states that if the Elliot-Halberstam Conjecture is true, then there are infinitely many consecutive primes differing by less than 16.

I can't seem to find any similar results that don't depend on unproven conjectures, so proof might not be easy.

A proven result that might be useful is that [itex]{\lim \inf}_{n \to \infty} \frac{p_{n+1}-p_n}{\log{p_n}}=0[/itex] (found in the same paper).

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- #19

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I'll try and figure it out myself. If I get stuck maybe I'll crack and ask for your proof.

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If you do manage to prove it, I'd love to see your proof. Please do post it.

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- #22

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From Wikipedia:

[tex] 0.922 \frac{n}{\ln n} < \pi(n) < 1.105 \frac{n}{\ln n}.[/tex]

That looks like enough to me.

Edit: Woops! Looks like I'm a little late. I hate posting in older threads by accident!

From Mathworld you'll also find a reference showing thatIn 1940, Paul Erdős showed that there is a constant c < 1 and infinitely many primes p such that (p′ - p) < (c ln p) where p′ denotes the next prime after p. This result was successively improved; in 1986 Helmut Maier showed that a constant c < 0.25 can be used. In 2004 Daniel Goldston and Cem Yıldırım showed that the constant could be improved further to c = 0.085786… In 2005, Goldston, János Pintz and Yıldırım established that c can be chosen arbitrarily small [1], [2]:

[tex] 0.922 \frac{n}{\ln n} < \pi(n) < 1.105 \frac{n}{\ln n}.[/tex]

That looks like enough to me.

Edit: Woops! Looks like I'm a little late. I hate posting in older threads by accident!

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- #23

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My proof is as follows:

We want to show, that for any [itex]m[/itex] no matter how large, there is an [itex]n>m[/itex] such that

[tex]\frac{n}{p_n}<\frac{n+1}{p_{n+1}}[/tex]

We can rearrange this:

[tex]np_{n+1}<p_n(n+1)[/tex]

[tex]np_{n+1}-np_n<p_n[/tex]

[tex]p_{n+1}-p_n<\frac{p_n}{n}[/tex]

[tex]\frac{p_{n+1}-p_n}{\log{p_n}}<\frac{p_n}{n\log{p_n}}[/tex]

Thus our theorem is established if we can show that [itex]\frac{p_{n+1}-p_n}{\log{p_n}}[/itex] can be made smaller than [itex]\frac{p_n}{n\log{p_n}}[/itex] for [itex]n[/itex] arbitrarily large.

It is an established result by Daniel Goldston and Cem Yıldırım that

[tex]{\lim \inf}_{n \to \infty} \frac{p_{n+1}-p_n}{\log{p_n}}=0[/tex]

Therefore, for any [itex]\epsilon>0[/itex], we can find an arbitrarily large [itex]n[/itex] such that

[tex]\frac{p_{n+1}-p_n}{\log{p_n}}<\epsilon[/tex]

The prime number theorem states that

[tex]\pi(n) \sim \frac{n}{\log{n}}[/tex]

which is useful here:

[tex]n = \pi(p_n) \sim \frac{p_n}{\log{p_n}}[/tex]

This means concretely that

[tex]\lim_{n \to \infty} \frac{p_n}{n\log{p_n}} = 1[/tex]

Therefore we can find an [itex]N[/itex] such that for all [itex]n \geq N[/itex],

[tex]\frac{p_n}{n\log{p_n}} > \frac{1}{2}[/tex]

But by Goldston and Yıldırım, we can find an [itex]n[/itex] as large as we like such that

[tex]\frac{p_{n+1}-p_n}{\log{p_n}}<\frac{1}{2}[/tex]

and thus we have established our theorem!

---Q-E-D---

If you have any questions about it, please post them. :)

We want to show, that for any [itex]m[/itex] no matter how large, there is an [itex]n>m[/itex] such that

[tex]\frac{n}{p_n}<\frac{n+1}{p_{n+1}}[/tex]

We can rearrange this:

[tex]np_{n+1}<p_n(n+1)[/tex]

[tex]np_{n+1}-np_n<p_n[/tex]

[tex]p_{n+1}-p_n<\frac{p_n}{n}[/tex]

[tex]\frac{p_{n+1}-p_n}{\log{p_n}}<\frac{p_n}{n\log{p_n}}[/tex]

Thus our theorem is established if we can show that [itex]\frac{p_{n+1}-p_n}{\log{p_n}}[/itex] can be made smaller than [itex]\frac{p_n}{n\log{p_n}}[/itex] for [itex]n[/itex] arbitrarily large.

It is an established result by Daniel Goldston and Cem Yıldırım that

[tex]{\lim \inf}_{n \to \infty} \frac{p_{n+1}-p_n}{\log{p_n}}=0[/tex]

Therefore, for any [itex]\epsilon>0[/itex], we can find an arbitrarily large [itex]n[/itex] such that

[tex]\frac{p_{n+1}-p_n}{\log{p_n}}<\epsilon[/tex]

The prime number theorem states that

[tex]\pi(n) \sim \frac{n}{\log{n}}[/tex]

which is useful here:

[tex]n = \pi(p_n) \sim \frac{p_n}{\log{p_n}}[/tex]

This means concretely that

[tex]\lim_{n \to \infty} \frac{p_n}{n\log{p_n}} = 1[/tex]

Therefore we can find an [itex]N[/itex] such that for all [itex]n \geq N[/itex],

[tex]\frac{p_n}{n\log{p_n}} > \frac{1}{2}[/tex]

But by Goldston and Yıldırım, we can find an [itex]n[/itex] as large as we like such that

[tex]\frac{p_{n+1}-p_n}{\log{p_n}}<\frac{1}{2}[/tex]

and thus we have established our theorem!

---Q-E-D---

If you have any questions about it, please post them. :)

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- #24

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An alternate argument using what I posted (when you add the mathworld result, I'm using on the one hand stronger and on the other hand weaker results than Moo did in his proof. They're almost identical. I'd say his is cleaner, though):

We have (from the mathworld ref.), for all natural n,

[tex]0.5 \frac{p_n}{\ln p_n} < \pi(p_n) = n < 2\frac{p_n}{\ln p_n},[/tex]

so in particular

[tex]\frac{p_n}{n} > \frac{\ln p_n}{2}[/tex].

Let N be some natural. By my Wikipedia ref., there is [itex]c<1/2[/itex] and [itex]n>N[/itex] s.t.

[tex]p_{n+1} - p_n < c\ln p_n < \frac{\ln p_n}{2} < \frac{p_n}{n},[/tex]

as required.

We have (from the mathworld ref.), for all natural n,

[tex]0.5 \frac{p_n}{\ln p_n} < \pi(p_n) = n < 2\frac{p_n}{\ln p_n},[/tex]

so in particular

[tex]\frac{p_n}{n} > \frac{\ln p_n}{2}[/tex].

Let N be some natural. By my Wikipedia ref., there is [itex]c<1/2[/itex] and [itex]n>N[/itex] s.t.

[tex]p_{n+1} - p_n < c\ln p_n < \frac{\ln p_n}{2} < \frac{p_n}{n},[/tex]

as required.

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- #25

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Fascinating. Thanks to you both!

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