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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 $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} \ge 1$ and a second argument covers the condition where $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} < 1$

I am very glad to be corrected if someone finds 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:

$$\ln\Gamma(\frac{x}{b_1}) - \ln\Gamma(\frac{x}{b_2} + \frac{1}{2}) - \ln\Gamma(\frac{x}{b_3} + \frac{1}{2})\le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$

when:

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

and:

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

Here's the argument for this generalization:

Let:

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

where:

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

Since:

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

We have:

$$\{\frac{x}{b_1}\} + 1 = \{\frac{x}{b_2}\} + \{\frac{x}{b_3}\}$$

So that:

$$-\{\frac{x}{b_1}\} - 1 = -\{\frac{x}{b_2}\} + -\{\frac{x}{b_3}\}$$

$$1-\{\frac{x}{b_1}\} = 1-\{\frac{x}{b_2}\} + 1-\{\frac{x}{b_3}\}$$

$$-\{\frac{x}{b_1}\} = 1-\{\frac{x}{b_2}\} - \frac{1}{2} + 1-\{\frac{x}{b_3} \} - \frac{1}{2}$$

$$\lfloor\frac{x}{b_1}\rfloor - \frac{x}{b_1} = (\lfloor\frac{x}{b_2}\rfloor + 1 - \frac{x}{b_2} - \frac{1}{2}) + (\lfloor\frac{x}{b_3}\rfloor + 1 - \frac{x}{b_3} - \frac{1}{2})$$

$$\lfloor\frac{x}{b_1}\rfloor + 1 = (\lfloor\frac{x}{b_2}\rfloor + 1) + (\lfloor\frac{x}{b_3}\rfloor + 1)$$

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:

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

Let:

$x_1 = \frac{x}{b_1}$, $\Delta{t_1} = 1 - \{\frac{x}{b_1}\}$,

$x_2 = \frac{x}{b_2}+\frac{1}{2}$, $\Delta{t_2} = \frac{1}{2} - \{\frac{x}{b_2}\}$

$x_3 = \frac{x}{b_3}+\frac{1}{2}$, $\Delta{t_3} = \frac{1}{2} - \{\frac{x}{b_3}\}$

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

Then:

$$\frac{\Gamma(\lfloor\frac{x}{b_1}\rfloor+1)}{\Gamma(\frac{x}{b_1})} \ge \frac{\Gamma(\lfloor\frac{x}{b_2}\rfloor+1)}{\Gamma(\frac{x}{b_2} + \frac{1}{2})}\frac{\Gamma(\lfloor\frac{x}{b_3}\rfloor+1)}{\Gamma(\frac{x}{b_3}+\frac{1}{2})}$$

So then it follows:

$$\ln\Gamma(\lfloor\frac{x}{b_1}\rfloor + 1) - \ln\Gamma(\frac{x}{b_1}) \ge \ln\Gamma(\lfloor\frac{x}{b_2}\rfloor + 1) - \ln\Gamma(\frac{x}{b_2} + \frac{1}{2}) + \ln\Gamma(\lfloor\frac{x}{b_3}\rfloor + 1) - \ln\Gamma(\frac{x}{b_3} + \frac{1}{2})$$

And we have shown:

$$\ln\Gamma(\frac{x}{b_1}) - \ln\Gamma(\frac{x}{b_2}+\frac{1}{2}) - \ln\Gamma(\frac{x }{b_3}+\frac{1}{2}) \le \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$

Here's an example:

Let $x=78.9$, $b_1=6$, $b_2=7$, $b_3=42$

So that:

$\{\frac{78.9}{7}\} = 0.15$, $\{\frac{78.9}{42}\} \approx 0.8786$

Then:

$$\ln\Gamma(\frac{78.9}{6}) - \ln\Gamma(\frac{78.9}{7}+\frac{1}{2}) - \ln\Gamma(\frac{78.9}{42} + \frac{1}{2}) \le \ln(\lfloor\frac{78.9}{6}\rfloor!) - \ln(\lfloor\frac{78.9}{7}\rfloor!) - \ln(\lfloor\frac{78.9}{42}\rfloor!)$$

Please let me know if you see any mistakes.

Thanks,

-Larry

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