# Generalizing Ramanujan's proof of Bertrand's Postulate: Can Ramanujan's approach be used to show a prime between $4x$ and $5x$ for $x \ge 3$

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)]:

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

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

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?
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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 at 13:59
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 at 13:40
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 at 16:21
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 at 16:52
@Douglas, thanks for mention about Ribenboim. I'll check it out this evening. Cheers. –  Larry Freeman Apr 16 at 18:49