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Perhaps, I've been thinking too long about Ramanujan's proof, but it appears to me that his argument can be generalized beyond $x$ and $2x$. My argument below attempts to show that for $x \ge 1331$, there is always a prime between $4x$ and $5x$.

I can use a similar argument to establish there is a prime between $2x$ and $3x$ and between $3x$ and $4x$. Based on some rough estimates, it looks it should also work to prove a prime between $5x$ and $6x$ as well as a prime between $6x$ and $7x$.

Since I am still getting up to speed on analytic number theory, I will be very glad if someone can point out the mistake that I am making in my reasoning. I am not yet able to find it.

Let $$\vartheta(x) = \sum_{p \le x}\ln(p)$$

Let $$\psi(x) = \sum_{n=1}^{\infty}\vartheta(x^{\frac{1}{n}})$$

Following Ramanujan [see (6)]:

$$\psi(x) - 2\psi(\sqrt{x}) \le \vartheta(x) \le \psi(x)$$

Analogous to Ramanujan's statement about:

$$\ln(\lfloor{x}\rfloor]!) - \ln(\lfloor\frac{x}{2}\rfloor!) - \ln(\lfloor\frac{x}{2}\rfloor!) = \psi(x) - \psi(\frac{x}{2}) + \psi(\frac{x}{3}) - \psi(\frac{x}{4}) + \ldots$$

Here's my restatement in terms of $4x$ and $5x$:

$$\ln(\lfloor\frac{x}{4}\rfloor!) - \ln(\lfloor\frac{x}{5}\rfloor!) - \ln(\lfloor\frac{x}{20}\rfloor!) = \psi(\frac{x}{4}) - \psi(\frac{x}{5}) + \psi(\frac{x}{8}) - \psi(\frac{x}{10}) + \ldots$$

where for each successive term we can see:

$$\psi(\frac{x}{4}) \ge \psi(\frac{x}{5}) \ge \psi(\frac{x}{8}) \ge \psi(\frac{x}{10}) \ge \ldots$$

Since, for any integer $v \ge 1$, we have:

$$\psi(\frac{x}{20v+4}) - \psi(\frac{x}{20v+5}) + \psi(\frac{x}{20v+8})-\psi(\frac{x}{20v+10})+\psi(\frac{x}{20v+12}) -\psi(\frac{x}{20v+15}) + \psi(\frac{x}{20v+16}) - \psi(\frac{x}{20v+20}) + \ldots$$

That is, a decreasing sequence of real numbers tending to 0, where each successive term has an alternating sign.

So, based on reasoning found here, it follows:

$$\psi(\frac{x}{4}) - \psi(\frac{x}{5}) + \psi(\frac{x}{8}) - \psi(\frac{x}{10}) + \psi(\frac{x}{12}) \ge \ln(\lfloor\frac{x}{4}\rfloor!) - \ln(\lfloor\frac{x}{5}\rfloor!) -\ln(\lfloor\frac{x}{20}\rfloor!)$$

From $\psi(x) - 2\psi(\sqrt{x}) \le \vartheta(x) \le \psi(x)$, it follows that:

$$\psi(\frac{x}{4}) - \psi(\frac{x}{5}) + \psi(\frac{x}{8}) - \psi(\frac{x}{10}) + \psi(\frac{x}{12}) \le \vartheta(\frac{x}{4}) - \vartheta(\frac{x}{5}) + 2\psi(\sqrt{\frac{x}{4}}) + \psi(\frac{x}{8}) - \psi(\frac{x}{10}) + \psi(\frac{x}{12})$$

Using the same reasoning as above, it can be noted that:

$$\psi(\frac{x}{10}) - \psi(\frac{x}{12}) \le \ln(\lfloor\frac{x}{10}\rfloor!) - \ln(\lfloor\frac{x}{12}\rfloor!) - \ln(\lfloor\frac{x}{60}\rfloor!)$$

So that we have:

$$\psi(\frac{x}{4}) - \psi(\frac{x}{5}) + \psi(\frac{x}{8}) - \psi(\frac{x}{10}) + \psi(\frac{x}{12}) \le \vartheta(\frac{x}{4}) - \vartheta(\frac{x}{5}) + 2\psi(\sqrt{\frac{x}{4}}) + \psi(\frac{x}{8}) - [ \ln(\lfloor\frac{x}{10}\rfloor!) - \ln(\lfloor\frac{x}{12}\rfloor!) - \ln(\lfloor\frac{x}{60}\rfloor!) ]$$

which implies:

$$\vartheta(\frac{x}{4}) - \vartheta(\frac{x}{5}) \ge \ln(\lfloor\frac{x}{4}\rfloor!) - \ln(\lfloor\frac{x}{5}\rfloor!) -\ln(\lfloor\frac{x}{20}\rfloor!) - 2\psi(\sqrt{\frac{x}{4}}) - \psi(\frac{x}{8}) + \ln(\lfloor\frac{x}{10}\rfloor!) - \ln(\lfloor\frac{x}{12}\rfloor!) - \ln(\lfloor\frac{x}{60}\rfloor!)$$

From Rosser and Schoenfeld (1961), we know that (see Theorem 12):

$$\psi(x) < 1.03883x$$

So that:

$$\vartheta(\frac{x}{4}) - \vartheta(\frac{x}{5}) \ge \ln(\lfloor\frac{x}{4}\rfloor!) - \ln(\lfloor\frac{x}{5}\rfloor!) -\ln(\lfloor\frac{x}{20}\rfloor!) - 2(1.03883)(\sqrt{\frac{x}{4}}) - (1.03883)(\frac{x}{8}) + \ln(\lfloor\frac{x}{10}\rfloor!) - \ln(\lfloor\frac{x}{12}\rfloor!) - \ln(\lfloor\frac{x}{60}\rfloor!)$$

Based on Stirling's Approximation and my reasoning found here, it follows that $\vartheta(\frac{x}{4}) - \vartheta(\frac{x}{5}) > 0$ for $x \ge 1331$

I have also verified that for $1331 > x > 2$, there is always a prime between $5x$ and $4x$ so if my argument is valid, this would be enough to establish that there is always a prime between $5x$ and $4x$ for $x \ge 3$.

Is this approach valid?

Update: I have found my mistake. The following step is invalid:

$$\psi(\frac{x}{4}) - \psi(\frac{x}{5}) + \psi(\frac{x}{8}) - \psi(\frac{x}{10}) + \psi(\frac{x}{12}) \le \vartheta(\frac{x}{4}) - \vartheta(\frac{x}{5}) + 2\psi(\sqrt{\frac{x}{4}}) + \psi(\frac{x}{8}) - [ \ln(\lfloor\frac{x}{10}\rfloor!) - \ln(\lfloor\frac{x}{12}\rfloor!) - \ln(\lfloor\frac{x}{60}\rfloor!) ]$$

Edit: I have added a clarification below on what type of answer I am looking for to this question.

Clarification: I am especially interested in one of these answers to this question:

  • Is this approach already "well-understood" (in which case, I would be interested in a reference)
  • Does this approach have "a major gap" (if so, which part of the argument is wrong or needs additional detail)
  • Could it be interesting "if it shows..." (what result is needed for this approach to be interesting to mathematician).
  • How could it be "improved and made more clear..." (what theorems or analytic techniques would strengthen or clarify the argument)
  • If the argument looks good, what would be the recommended next step from here?
Glorfindel
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Larry Freeman
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  • The formula you give after "Since, for any integer v≥1, we have:" should have inequality sign $\geq 0$. – Sungjin Kim Apr 10 '13 at 08:31
  • Thanks for your comment. I am trying to show that that this is a decreasing sequence of real numbers with an alternating sign. I'll update to make my point clearer. – Larry Freeman Apr 10 '13 at 08:38
  • Perhaps you should elaborate on "Specifically, I note that:" this part. – Sungjin Kim Apr 10 '13 at 08:49
  • Sure. I'll update it now. I'll try to make my point clearer. – Larry Freeman Apr 10 '13 at 08:52
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    Note that you can get displayed equations by enclosing them in double dollar signs. A lot of these equations would be easier to read that way (and it looks nicer when they're centred). – joriki Apr 11 '13 at 13:59
  • Thanks for the suggestion. I did as you requested. Please let me know if there are any more opportunities to make the question more readable. – Larry Freeman Apr 11 '13 at 17:54
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    You might look at the work of Bachraoui and Nagura if you haven't already. We also have a general proof that there is a prime on $(x,(1+1/k)x)$ for large enough x. – daniel Apr 14 '13 at 13:40
  • Thanks, Daniel! I have looked at Nagura and Bachraoui. Bachraoui seems very much influenced by Paul Erdős's take on Bertrand's Posulate. I would be very interested in the general proof of a prime on $(x,(1+\frac{1}{k})x)$. Do you have the reference? I found this amazing link via Terence Tao that talks about the possibility of such a proof. Here's the discussion. – Larry Freeman Apr 15 '13 at 03:04
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    The fact that there is a prime in the interval $(x, (1+\frac{1}{k})x)\ \forall k\in \mathbb{N}, \forall x\geq x_k$ for some $x_k\in \mathbb{N}$ follows directly from the prime number theorem; this is discussed on pp. 397-398 of Paulo Ribenboim's "The New Book of Prime Number Records". – Douglas B. Staple Apr 16 '13 at 16:21
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    Ribenboim also points out in that reference that there is a prime $p$ with $kn < p < (k+1)n\ \forall n\geq 9$ and $0<k\leq 7$ as a corollary to this 1932 proof by Breusch. – Douglas B. Staple Apr 16 '13 at 16:52
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    @Douglas, thanks for mention about Ribenboim. I'll check it out this evening. Cheers. – Larry Freeman Apr 16 '13 at 18:49
  • Hi @Douglas. I'm very glad. Thanks for your help. – Larry Freeman Apr 16 '13 at 21:25
  • Hi Doug, I'm most interested in the correctness of the argument. Assuming it is valid (which I'm not yet convinced), I would then be interested if there's anything interesting in the approach? Is the result already well-known? – Larry Freeman Apr 17 '13 at 22:31
  • By proving $\displaystyle\lim_{n\to\infty}\frac{p_n}{p_{n-1}}=1$. It can be even generalised that there is a prime between $x$ and $x(1+\epsilon)$ for large enough $x$. – Grobber Apr 18 '13 at 10:33
  • Hi Douglas, to be clear, is the same result established with an effective bound and with the same reasoning? If it's derived from the Prime Number Theorem, as I understand it from here, that's not the same result. I tried to read the Breusch paper but it is in German. I looked through the Ribenboim page you cited on p397-398 of NB of PNR but in my version of the book, that page is a section is Chapter 6, II, "Prime Values of Polynomials of Arbitrary Degree." – Larry Freeman Apr 18 '13 at 14:12
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    Hey Larry, the result is the same but the reasoning is not. I probably should've mentioned that the Breusch paper is in German; see the description in Ribenboim's book instead. My copy of Ribenboim's book is the 3rd Ed.; I think you have the correct part of book. It starts with "in the same paper, there is the following conjecture by Sierpinski..." and then goes on to discuss partial proofs of the conjecture, where the Breusch paper is discussed. Search Ribenboim's index for the Breusch paper. – Douglas B. Staple Apr 18 '13 at 15:05
  • Douglas, thanks for the clarifications. – Larry Freeman Apr 18 '13 at 22:43
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    Perhaps you would be interested in the following paper: Meher, J., & Murty, M. R. (2013). Ramanujan's Proof of Bertrand's Postulate. American Mathematical Monthly, 120(7), 650-653. (Note that the necessity of Stirling's formula is eliminated in this paper.) – Benjamin Dickman Dec 02 '13 at 08:19
  • Thanks, @Benjamin, sounds very interesting! I will take a look. :-) – Larry Freeman Dec 02 '13 at 16:49
  • Just looking through the badges, congrats on a year! –  Nov 08 '15 at 06:04

2 Answers2

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Ramanujan's proof is actually a simplification of Chebyshev's original [1852] proof of Bertrands postulate (the article is Memoire sur les nombres premiers. J. Math. Pures Appl. 17 1852). Chebyshev uses a stronger approach proving a lot more than Bertrand's postulate, in particular your statement follows directly from his bound while his methods are essentially the same as Ramanujan's

He starts deriving the identity $$ \log [x]! = \psi(x) + \psi\left(\frac x2 \right) + \psi\left(\frac x3 \right) + \dots $$ and then he uses it to derive the following identity (similar to Ramanujan's or your's but stronger):

$$ \log \frac{ \left\lfloor x\right\rfloor! \left\lfloor \frac x{30} \right\rfloor!}{ \left\lfloor \frac x2 \right\rfloor!\left\lfloor \frac x3 \right\rfloor!\left\lfloor \frac x5 \right\rfloor! } = \psi\left(x\right)-\psi\left(\frac{x}{6}\right)+\psi\left(\frac{x}{7}\right) -\psi\left(\frac{x}{10}\right)+\psi\left(\frac{x}{11}\right) -\psi\left(\frac{x}{12}\right)+\psi\left(\frac{x}{13}\right) -\psi\left(\frac{x}{15}\right)+\psi\left(\frac{x}{17}\right) -\psi\left(\frac{x}{18}\right)+\psi\left(\frac{x}{19}\right) -\psi\left(\frac{x}{20}\right)+\psi\left(\frac{x}{23}\right) -\psi\left(\frac{x}{24}\right)+\psi\left(\frac{x}{29}\right) -\psi\left(\frac{x}{30}\right) + \dots - \dots $$ where the sequence in the right continues with period 30 in he denominator, ie the first missing terms are $\psi(x/31)-\psi(x/36)+\psi(x/37)- \dots $.

As you can see this is similar to the argument you give above, as we have an alternating sequence of non-increasing terms, so you can bound it above and below stopping after an odd/even number of terms: $$ \psi(x) - \psi\left(\frac x6\right) \le \log \frac{ \left\lfloor x\right\rfloor! \left\lfloor \frac x{30} \right\rfloor!}{ \left\lfloor \frac x2 \right\rfloor!\left\lfloor \frac x3 \right\rfloor!\left\lfloor \frac x5 \right\rfloor! } \le \psi(x) $$

Now he uses Stirling to find the approximation $$ Ax - \frac{5}{2}\log x - 1 < \log \frac{ \left\lfloor x\right\rfloor! \left\lfloor \frac x{30} \right\rfloor!}{ \left\lfloor \frac x2 \right\rfloor!\left\lfloor \frac x3 \right\rfloor!\left\lfloor \frac x5 \right\rfloor! } < A x + \frac{5}{2}\log x $$ where $$ A = \log \frac{2^{1/2}3^{1/3}5^{1/5}}{30^{1/30}} = 0.92129202 $$ which combined with the previous inequalities gives: $$ \psi(x) > Ax -\frac{5}{2}\log x -1 \quad\text{and}\quad \psi(x)-\psi\left(\frac{x}{6}\right) < Ax + \frac{5}{2}\log x $$ He uses an auxiliary funciont to telescope this inequality and finds that: $$ \psi(x) < \frac{6}{5}Ax + \frac{5}{4\log 6} \log^2 x + \frac{5}{4}\log x + 1 $$ And now he uses the inequality $$ \psi(x) - 2\psi(\sqrt{x}) < \theta(x) < \psi(x) - \psi(\sqrt{x}) $$ to derive (after some technical work) that there are more than $k$ primes between $l$ and $L$ if $$ l = \frac{5}{6} L - 2 \sqrt{L} - \frac{25 \log^2 L}{16\log 6 A}-\frac{5}{6A}\left(\frac{25}{4}+k\right)\log L - \frac{25}{6A} $$ So with $k = 0$ we find that there is a prime between $4x$ and $5x$ for $x >= 2034$ and we can check numerically that the same holds for $x \ge 1331$.

(However this falls short to prove that there is always a prime between $5x$ and $6x$.)

Some comments about your clarifications:

  • As it has been commented there is an elementary proof of the prime number theorem so there is an elementary proof that there is a prime between $x$ and $(1 + \epsilon) x$ for all $\epsilon$.

  • This is the first result of his kind, but the bounds obtained by Chebyshev are not easy to improve without using the PNT. First Sylvester [1881] On Tchebycheff's theory of the totality of the prime numbers comprised within given limits. Improved the upper bound on $\psi(x)$ to $$ 0.95695 x \le \psi(x) \le 1.04423 x $$ Using (I think) combinations of identities as Chebyshev's and Ramanujan's. I never was able to find the article so I'm not sure about Sylvester's method or results (so please check them). With these inequalities it should be possible to prove the existence of a prime between $10x$ and $11x$ for large enough $x$.

  • In http://arxiv.org/pdf/0709.1977v1.pdf you can find a list of all possible identities that can be used almost directly in Chebyshev's method, there are several infinite families and 52 isolated identities. (however I believe Chebyshev's identity is the one that gives the best possible bound using only his method).

  • I think using Rosser and Schoenfeld's result $\psi(x)<1.03883x$ is (in some sense) a major gap in your argument. I don't know how it is derived but I think as there are a lot of results in that article that imply the existence of a prime between $kx$ and $(k+1)x$ for every prime, so I would avoid it to prevent circular reasoning.

Woett
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Esteban Crespi
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0

See and search for Generalized Ramanujan Primes

https://en.wikipedia.org/wiki/Ramanujan_prime#Generalized_Ramanujan_primes

John Nicholson
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