I am investigating the generality of Lemma 1 in Nagura's proof that there is always a prime between $x$ and $\frac{6x}{5}$.

In Lemma 1, Nagura establishes when $n > 1$, $x \ge 1$:

$$\frac{1}{n}\log\left(\left\lfloor{x}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}\right\rfloor!\right) \ge \frac{1}{n}\log\Gamma\left(x\right) - \log\Gamma\left(\frac{x+n-1}{n}\right)$$

and for $n > 1$, $x \ge n$:

$$\frac{1}{n}\log\left(\left\lfloor{x}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}\right\rfloor!\right) \le \frac{1}{n}\log\Gamma\left(x+1\right) - \log\Gamma\left(\frac{x+1}{n}\right)$$

I am trying to establish a similar inequality for $\frac{2}{n}\log\left(\left\lfloor\frac{x}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}\right\rfloor!\right)$

Here's what I am trying to prove:

If $\left\{\frac{x}{2}\right\} \ge \frac{1}{2} + \frac{\left\{x\right\}}{2}$, $n > 2$, $x \ge n$, then:

$$\frac{2}{n}\log\left(\left\lfloor\frac{x}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}\right\rfloor!\right) \le \frac{2}{n}\log\Gamma\left(\frac{x+1}{2}\right) - \log\Gamma\left(\frac{x+1}{n}\right)$$

Here's the argument:

Let $\left\{k\right\}$ be the fractional part of $k$ so that $\left\{k\right\} = k - \left\lfloor{k}\right\rfloor$.

Since $\frac{\left\lfloor{x}\right\rfloor+1}{n} \le \left\lfloor\frac{x}{n}\right\rfloor+1$ (see answer here),

$$\frac{2}{n}\log\left(\left\lfloor\frac{x}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}\right\rfloor!\right) \le \frac{2}{n}\log\Gamma\left(\frac{x}{2} - \{\frac{x}{2}\} + 1\right) - \log\Gamma\left(\frac{\left\lfloor{x}\right\rfloor+1}{n}\right)$$

Since $\left\{\frac{x}{2}\right\} \ge \frac{1}{2} + \frac{\left\{x\right\}}{2}$, it follows:

$$\frac{2}{n}\log\Gamma\left(\frac{x}{2} - \{\frac{x}{2}\} + 1\right) - \log\Gamma\left(\frac{\left\lfloor{x}\right\rfloor+1}{n}\right) = \frac{2}{n}\log\Gamma\left(\frac{\left\lfloor{x}\right\rfloor}{2} + \frac{\left\{x\right\}}{2} - \left\{\frac{x}{2}\right\} + 1\right) - \log\Gamma\left(\frac{\left\lfloor{x}\right\rfloor+1}{n}\right) \le \frac{2}{n}\log\Gamma\left(\frac{\left\lfloor{x}\right\rfloor}{2} + \frac{1}{2}\right) - \log\Gamma\left(\frac{\left\lfloor{x}\right\rfloor+1}{n}\right) = \frac{2}{n}\log\Gamma\left(\frac{\left\lfloor{x}\right\rfloor+1}{2} \right) - \log\Gamma\left(\frac{\left\lfloor{x}\right\rfloor+1}{n}\right)$$

To complete the argument, I need to show that $\frac{2}{n}\log\Gamma\left(\frac{x+1}{2}\right) - \log\Gamma\left(\frac{x+1}{n}\right)$ is an increasing function.

Since $n > 2$, it follows that:

$\frac{x+1}{2} > \frac{x+1}{n}$

Using $\frac{\Gamma'}{\Gamma}\left(s\right) =\int_0^{\infty}\left(\frac{e^{-t}}{t}-\frac{e^{-st}}{1 - e^{-t}}\right)dt$, when $s > 0$, we have for ($x > 0$):

$$\frac{\Gamma'}{\Gamma}\left(\frac{x+1}{2}\right) - \frac{\Gamma'}{\Gamma}\left(\frac{x+1}{n}\right) = \int_0^{\infty}\frac{1}{1 - e^{-t}}\left(e^{-\frac{x+1}{n}t} - e^{-\frac{x+1}{2}t}\right)dt > 0$$

Since $\frac{2}{n}\log\Gamma\left(\frac{x+1}{2}\right) - \log\Gamma\left(\frac{x+1}{n}\right)$ is an increasing function, it follows:

$$\frac{2}{n}\log\Gamma\left(\frac{\left\lfloor{x}\right\rfloor+1}{2} \right) - \log\Gamma\left(\frac{\left\lfloor{x}\right\rfloor+1}{n}\right) \le \frac{2}{n}\log\Gamma\left(\frac{x+1}{2}\right) - \log\Gamma\left(\frac{x+1}{n}\right)$$

Does this work? Have I made a mistake? If I haven't made a mistake is there an easier way to prove it?



  • 1
    $\begingroup$ Is the condition $\{\frac{x}{2}\} \ge \frac{1}{2} + \frac{\lfloor{x}\rfloor}{2}$ correct? This is true only when $x < 0$, and then the left-hand side of the inequality you want to prove is undefined. Your inequality does not hold for all $x > 0$, for example when $n=3$ and $x=4$. $\endgroup$ – Antonio Vargas May 4 '13 at 1:47
  • $\begingroup$ Thanks for catching that! The condition should have read: $\{\frac{x}{2}\} \ge \frac{1}{2} + \frac{\{x\}}{2}$. I've updated my question. $\endgroup$ – Larry Freeman May 4 '13 at 2:28
  • 1
    $\begingroup$ It seems that the condition on $x$ must also have some kind of dependence on $n$, since the inequality is also false for $n=10$, $x=7.5$. $\endgroup$ – Antonio Vargas May 4 '13 at 2:57
  • $\begingroup$ Very good point. It appears that there is a problem if $x < n$. In the case of $n=10$, $x=7.5$, the issue seems to be that $\log\Gamma(0.8) > \log\Gamma(1)$. I've updated my condition to include $x \ge n$. $\endgroup$ – Larry Freeman May 4 '13 at 3:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.