# Is it possible to generalize Ramanujan's lower bound for factorials that he used in his proof of Bertrand's Postulate?

I am starting to feel more confident in my understanding of Ramanujan's proof of Bertrand's postulate. I hope that I am not getting overconfident.

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:

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

I would really appreciate it if my argument could be reviewed and someone could call out any mistakes either in the answer or in comments. :-)

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:

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

So that:

$$-\{\frac{x}{b_1}\} \ge -\{\frac{x}{b_2}\} + -\{\frac{x}{b_3}\}$$

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

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

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

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

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:

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

Please let me know if you see any mistakes.

Thanks,

-Larry

Edit: Based on reviewing Zander's answer, I believe that this argument can be saved. The revision requires two separate arguments:

• one for: $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} \ge 1$
• another for: $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} < 1$

The link for the first argument is here. The link for the second argument is here.

After this step: $$\lfloor\frac{x}{b_1}\rfloor + 1 - \frac{x}{b_1} \ge (\lfloor\frac{x}{b_2}\rfloor + 1 - \frac{x}{b_2} - \frac{1}{2}) + (\lfloor\frac{x}{b_3}\rfloor + 1 - \frac{x}{b_3} - \frac{1}{2})$$ we have $$\lfloor\frac{x}{b_1}\rfloor + 1 \ge (\lfloor\frac{x}{b_2}\rfloor + 1) + (\lfloor\frac{x}{b_3}\rfloor + 1)-1$$ but what you have is incorrect. For example, let $x=77.1,b_1=6,b_2=7,b_3=42$, then $\lfloor x/b_1\rfloor=12, \lfloor x/b_2\rfloor=11, \lfloor x/b_3 \rfloor=1$ and $13\not\ge 14$.
And in this case your inequality is also violated: $$\ln \Gamma\left(\frac{77.1}{6}\right)-\ln\Gamma\left(\frac{77.1}{7}+\frac{1}{2}\right)-\ln \Gamma\left(\frac{77.1}{42}+\frac{1}{2}\right) = 3.10698\cdots \\ \ge \ln(12!)-\ln(11!)-\ln(1!) = 2.4849\cdots$$
• After reviewing the issue, I believe that the proof can be saved. I believe that the step in question is valid in the case where $\{\frac{x}{b_2}\} + \{\frac{x}{b_3}\} \ge 1$. Then, the step in question becomes $\lfloor\frac{x}{b_1}\rfloor - \frac{x}{b_1} = (\lfloor\frac{x}{b_2}\rfloor + 1 - \frac{x}{b_2} - {1}{2}) + (\lfloor\frac{x}{b_3}\rfloor + 1 - \frac{x}{b_3} - \frac{1}{2})$ which works. For example, let $x=78.9$ and $\lfloor\frac{x}{b_1}\rfloor+1\ge(\lfloor\frac{x}{b_2}\rfloor+1)+(\lfloor\frac{x}{b_3}\rfloor+1)$. To avoid confusion, I'll post a new question with a revised proof. – Larry Freeman Apr 14 '13 at 10:45