# Generalizing Jitsuro Nagura's result: my resulting upper bound for the second chebyshev function is too low. What am I doing wrong?

I've been reading through Jitsuro Nagura's classic proof that there is a prime between $x$ and $\frac{6x}{5}$ and it seems to me that it should be possible to improve on his upper bound for the second Chebyshev function.

In Nagura's paper, he establishes the following inequality:

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

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

Then, he establishes an upper bound using:

$$\psi\left(x\right) < 1.0851\left(x + \frac{x}{1806} + \frac{x}{1806^2} + \frac{x}{1806^3} + \ldots\right) < 1.086x$$

Using my analysis in this question and this question, and this effort to apply Stirling's formula in the same way as Nagura, then, for $x \ge 986$, I am finding:

$$T\left(\frac{x}{2}\right) - T\left(\frac{x}{3}\right) - T\left(\frac{x}{6}\right) < 0.321x$$

Applying the same approach as Nagura, I am finding:

$$T\left(\frac{x}{2}\right) - T\left(\frac{x}{3}\right) - T\left(\frac{x}{6}\right) \ge \psi\left(\frac{x}{2}\right) - \sum_{m=1}^{\infty}\left[\psi\left(\frac{x}{6m-3}\right) - \psi\left(\frac{x}{6m-2}\right) + \psi\left(\frac{x}{6m}\right) - \psi\left(\frac{x}{6m+2}\right) \right] \ge \psi\left(\frac{x}{2}\right) - \psi\left(\frac{x}{3}\right)$$

Putting it all together, I come up with:

$$\psi\left(\frac{x}{2}\right) < 0.321\left(x + \frac{x}{3} + \frac{x}{3^2} + \frac{x}{3^3} + \ldots\right) < \frac{3}{2}*0.321 = 0.4815x$$

Which, after seting $x = 2y$, results in:

$$\psi\left(y\right) < 0.4815\left(2y\right) = 0.963y$$

Considering the best known upper bound is $1.03883$, I am certainly doing something wrong.

Can anyone help me to figure out what's wrong with my analysis?

Thanks,

-Larry

Update: I wrote a simple java app to check $\psi(x)$. I am calculating $\psi(1627) > 1.01363*(1627)$. Hopefully, this will lead me to the mistake that I made. When I figure it out, I'll post it as part of this update.

So far, my suspicion is that the $0.321x$ is wrong. Based on $\psi(1627)$, it should be greater than $0.67575x$.

I found the mistake. :-)

This is wrong:

$$\psi\left(\frac{x}{2}\right) < 0.321\left(x + \frac{x}{3} + \frac{x}{3^2} + \frac{x}{3^3} + \ldots\right) < \frac{3}{2}*0.321 = 0.4815x$$

It should be:

$$\psi\left(\frac{x}{2}\right) < 0.321\left(x + \frac{2x}{3} + \frac{2^2{x}}{3^2} + \frac{2^3{x}}{3^3} + \ldots\right) < \frac{3}{1}*0.321 = 0.963x$$

• So, you have edited this question 13 times in a day. Just sayin'. – Gerry Myerson May 5 '13 at 7:34
• @GerryMyerson Isn't that admirable? He's doing the best to show his work. – Potato May 5 '13 at 9:06
• Hi Gerry, I've been editing it each time that I found a mistake or found an opportunity to make the display clearer (adding \left and \right) or make the details of the question clearer. – Larry Freeman May 5 '13 at 10:23
• Well, that's good, but maybe even better would be to take the time to get it right in the first place. – Gerry Myerson May 5 '13 at 12:17
• I agree, Gerry. It may not show but I believe that I am greatly improving as an amateur mathematician as a result of this great site. For the future, I'll do more evaluation upfront on paper so that my posts will require less edits and will hopefully be of superior quality. :-) – Larry Freeman May 5 '13 at 12:36