Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is the second attempt at a proof. My first attempt had a flaw in its logic.

After reviewing the mistake in logic, I believe that with a revised logic, the argument can be saved.

The revision consists of two arguments. The argument presented below covers the condition where $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} < 1$. My question here covers the condition where $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} \ge 1$

I am very glad to be corrected if someone is able to find a mistake. :-)

This is an attempt to generalize one of the steps in Ramanujan's proof of Bertrand's postulate.

In particular, Ramanujan's does the following comparison in step (8):

$$\ln\Gamma(x) - 2\ln\Gamma(\frac{x}{2} + \frac{1}{2}) \le \ln(\lfloor{x}\rfloor!) - 2\ln(\lfloor\frac{x}{2}\rfloor!)$$

It occurs to me that this can be generalized to:

$$\ln\Gamma(\frac{x}{b_1}-\frac{3}{16}) - \ln\Gamma(\frac{x}{b_2}+\frac{19}{32}) - \ln\Gamma(\frac{x}{b_3}+\frac{19}{32})\le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$


$$\frac{x}{b_1} = \frac{x}{b_2} + \frac{x}{b_3}$$


$$\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} < 1$$


$$x > 36$$

Here's the argument for this generalization:


$$\{\frac{x}{b_i}\} = \frac{x}{b_i} - \lfloor\frac{x}{b_i}\rfloor$$


$$0 \le \{\frac{x}{b_i}\} < 1$$


$$\{\frac{x}{b_1}\} + \lfloor\frac{x}{b_1}\rfloor = \{\frac{x}{b_2}\} + \lfloor\frac{x}{b_2}\rfloor + \{\frac{x}{b_3}\} + \lfloor\frac{x}{b_3}\rfloor$$

We have:

$$\{\frac{x}{b_1}\} = \{\frac{x}{b_2}\} + \{\frac{x}{b_3}\}$$


$$\lfloor\frac{x}{b_1}\rfloor = \lfloor\frac{x}{b_2}\rfloor + \lfloor\frac{x}{b_3}\rfloor$$

If $\Delta{t_1} \ge \Delta{t_2} + \Delta{t_3}$ and $x_1 + \Delta{t_1} \ge x_2 + \Delta{t_2} \ge x_3 + \Delta{t_3} > 0$,

Using the logic in the answer here:

$$\frac{\Gamma(x_1 + \Delta{t_1})}{\Gamma(x_1)} \ge \frac{\Gamma(x_2 + \Delta{t_2})}{\Gamma(x_2)}\frac{\Gamma(x_3 + \Delta{t_3})}{\Gamma(x_3)}$$


$x_1 = \frac{x}{b_1}-\frac{3}{16}$, $\Delta{t_1} = \frac{19}{16}-\{\frac{x}{b_1}\}$,

$x_2 = \frac{x}{b_2}+\frac{19}{32}$, $\Delta{t_2} = \frac{13}{32}-\{\frac{x}{b_2}\}$

$x_3 = \frac{x}{b_3}+\frac{19}{32}$, $\Delta{t_3} = \frac{13}{32}-\{\frac{x}{b_3}\}$

where $\frac{x}{b_2} \ge \frac{x}{b_3}$ (Otherwise, switch the two values).


$$\frac{\Gamma(\lfloor\frac{x}{b_1}\rfloor+1)}{\Gamma(\frac{x}{b_1}-\frac{3}{16})} \ge \frac{\Gamma(\lfloor\frac{x}{b_2}\rfloor+1)}{\Gamma(\frac{x}{b_2}+\frac{19}{32})}\frac{\Gamma(\lfloor\frac{x}{b_3}\rfloor+1)}{\Gamma(\frac{x}{b_3}+\frac{19}{32})}$$

So then it follows:

$$\ln\Gamma(\lfloor\frac{x}{b_1}\rfloor + 1) - \ln\Gamma(\frac{x}{b_1} - \frac{3}{16}) \ge \ln\Gamma(\lfloor\frac{x}{b_2}\rfloor+1) - \ln\Gamma(\frac{x}{b_2}+\frac{19}{32}) + \ln\Gamma(\lfloor\frac{x}{b_3}\rfloor + 1) - \ln\Gamma(\frac{x}{b_3}+\frac{19}{32})$$

And we have shown:

$$\ln\Gamma(\frac{x}{b_1}-\frac{3}{16}) - \ln\Gamma(\frac{x}{b_2}+\frac{19}{32}) - \ln\Gamma(\frac{x}{b_3}+\frac{19}{32}) \le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$

Here's an example:

Let $x=77.1$, $b_1=6$, $b_2=7$, $b_3=42$

So that:

$\{\frac{77.1}{7}\} \approx 0.0143$, $\{\frac{77.1}{42}\} \approx 0.8357$


$$\ln\Gamma(\frac{77.1}{6}-\frac{3}{16}) - \ln\Gamma(\frac{77.1}{7}+\frac{19}{32}) - \ln\Gamma(\frac{77.1}{42} + \frac{19}{32}) \le \ln(\lfloor\frac{77.1}{6}\rfloor!) - \ln(\lfloor\frac{77.1}{7}\rfloor!) - \ln(\lfloor\frac{77.1}{42}\rfloor!)$$

Please let me know if you see any mistakes.



Note 1: I came up with $\frac{x}{b_1}-\frac{3}{16}$ by comparing $\frac{x}{b_1}-1$ and then comparing $\frac{x}{b_1}-\frac{1}{2}$. I found that by taking consistent averages and satisfying these conditions:

If $\Delta{t_1} \ge \Delta{t_2} + \Delta{t_3}$ and $x_1 + \Delta{t_1} \ge x_2 + \Delta{t_2} \ge x_3 + \Delta{t_3} > 0$,

I was able to get a tighter lower bound. I should be able to improve upon $\frac{x}{b_1} - \frac{3}{16}$ if I wanted to.

Note 2: I have not been able to show that this tightening of the lower bound works in all cases. This is a gap in my argument. I still need to show that using $\frac{x}{b_1}-\frac{3}{16}$ is always an improvement to using the more obvious $\frac{x}{b_3}-1$. My argument shows that it will work but not that it is better.

share|cite|improve this question
up vote 1 down vote accepted

Looking at a graph it seems plausible that your result holds for $x>36$, but it does not hold for $35\le x <36$ where the right side is zero.

The flaw in this case is that $\{x/b_3\}>13/16$, allowing $\Delta t_3<0$ and $\Delta t_2>\Delta t_1$, violating the conditions for the $x_i+\Delta t_i$ inequality.

share|cite|improve this answer
Once again, thanks very much! Great point. I'll investigate the range of this result and update the question. I'll also work on evaluating when $\frac{x}{b_1} - \frac{3}{16}$ is a tighter lower bound than $\frac{x}{b_1} - 1$. – Larry Freeman Apr 14 '13 at 16:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.