Using Stirling's Formula to approximate a difference of logarithms of factorials in the same way as Jitsuro Nagura.

Question

In Jitsuro Nagura's classic proof of a prime existing between $x$ and $\frac{6x}{5}$, he uses Stirling's formula to show that:

$$T\left(x\right) - T\left(\frac{x}{2}\right) - T\left(\frac{x}{3}\right) - T\left(\frac{x}{7}\right) - T\left(\frac{x}{43}\right) - T\left(\frac{x}{1806}\right) < 1.0851x$$

where:

$$T\left(x\right) = \log\Gamma\left(\left\lfloor{x}\right\rfloor+1\right)$$

It seems to me that it should be possible to apply this same approach to:

$$T\left(\frac{x}{2}\right) - T\left(\frac{x}{3}\right) - T\left(\frac{x}{6}\right)$$

For purposes of my example, let's assume $\left\{\frac{x}{2}\right\} < \frac{1}{2} + \frac{\left\{x\right\}}{2}$, $n > 2$, $x \ge n$, $\exists{a} \in \mathbb{Z}, x < an \le \left(x+1\right)$:

Then, using my analysis in this question for left($x \ge 6$):

$$T\left(\frac{x}{2}\right) - T\left(\frac{x}{3}\right) - T\left(\frac{x}{6}\right) = \frac{2}{3}T\left(\frac{x}{2}\right) - T\left(\frac{x}{3}\right) + \frac{2}{6}T\left(\frac{x}{2}\right) - T\left(\frac{x}{6}\right) \le \log\Gamma\left(\frac{x+4}{2}\right) - \log\Gamma\left(\frac{x+4}{3}\right) - \log\Gamma\left(\frac{x+4}{6}\right)$$

Now, applying Stirling's formula of $\log\Gamma\left(x\right) = \left(x - \frac{1}{2} \right)\log{x}-x+\log\sqrt{2\pi}+\frac{\theta}{12x}$ with ($0 < \theta < 1$):

$$T\left(\frac{x}{2}\right) - T\left(\frac{x}{3}\right) - T\left(\frac{x}{6}\right) < \frac{x+4}{2}\log\frac{x+4}{2} - \frac{x+4}{3}\log\frac{x+4}{3} - \frac{x+4}{6}\log\frac{x+4}{6} - \frac{1}{2}\left(\log\frac{x+4}{2} - \log\frac{x+4}{3} - \log\frac{x+4}{6}\right) - \frac{x+4}{2} + \frac{x+4}{3} + \frac{x+4}{6} -\log\sqrt{2\pi} + \frac{1}{12\left(\frac{x+4}{2}\right)}$$

$$= \left(x+4\right)\left(-\frac{1}{2}\log{2} + \frac{1}{3}\log{3} + \frac{1}{6}\log{6}\right) + \frac{1}{2}\log\left(x+4\right) - \log{3} -\log\sqrt{2\pi} + \frac{1}{6\left(x+4\right)}$$

$$< 0.3182571x + \frac{1}{2}\log\left(x+4\right) - 0.744522 + \frac{1}{6\left(x+4\right)} < 0.321x$$

for $x \ge 986$

As a second example, I have found this inequality for the condition where $\left\{\frac{x}{2}\right\} \ge \frac{1}{2} + \frac{\left\{x\right\}}{2}$, $n > 2$, $x \ge n$

$$T\left(\frac{x}{2}\right) - T\left(\frac{x}{3}\right) - T\left(\frac{x}{6}\right) = \frac{2}{3}T\left(\frac{x}{2}\right) - T\left(\frac{x}{3}\right) + \frac{2}{6}T\left(\frac{x}{2}\right) - T\left(\frac{x}{6}\right) \le \log\Gamma\left(\frac{x+1}{2}\right) - \log\Gamma\left(\frac{x+1}{3}\right) - \log\Gamma\left(\frac{x+1}{6}\right)$$

which after the same analysis as above results in:

$$= \left(x+1\right)\left(-\frac{1}{2}\log{2} + \frac{1}{3}\log{3} + \frac{1}{6}\log{6}\right) + \frac{1}{2}\log\left(x+1\right) - \log{3} -\log\sqrt{2\pi} + \frac{1}{6\left(x+1\right)}$$

$$< 0.3182571x + \frac{1}{2}\log\left(x+1\right) - 1.699293 + \frac{1}{6\left(x+1\right)} < 0.321x$$

for $x \ge 522$

As a third example, I have found this inequality for the condition where $\left\{\frac{x}{2}\right\} < \frac{1}{2} + \frac{\left\{x\right\}}{2}$, $n > 2$, $x \ge n$, $\nexists{a} \in \mathbb{Z}, x < an \le \left(x+1\right)$

$$T\left(\frac{x}{2}\right) - T\left(\frac{x}{3}\right) - T\left(\frac{x}{6}\right) = \frac{2}{3}T\left(\frac{x}{2}\right) - T\left(\frac{x}{3}\right) + \frac{2}{6}T\left(\frac{x}{2}\right) - T\left(\frac{x}{6}\right) \le \log\Gamma\left(\frac{x+2}{2}\right) - \log\Gamma\left(\frac{x+2}{3}\right) - \log\Gamma\left(\frac{x+2}{6}\right)$$

which after the same analysis as above results in:

$$= \left(x+2\right)\left(-\frac{1}{2}\log{2} + \frac{1}{3}\log{3} + \frac{1}{6}\log{6}\right) + \frac{1}{2}\log\left(x+2\right) - \log{3} -\log\sqrt{2\pi} + \frac{1}{6\left(x+2\right)}$$

$$< 0.3182571x + \frac{1}{2}\log\left(x+2\right) - 1.38103665 + \frac{1}{6\left(x+2\right)} < 0.321x$$

for $x \ge 689$

Have I applied Stirling's formula correctly? Does anyone see any mistakes in this calculation?

Thanks very much,

-Larry