I've been told that the approach below will not work.

I would be interested if someone could help me to understand what will go wrong.


$$\psi(x) = \sum\limits_{p^k \le x} \ln p$$

So that (see here):

$$\ln(x!) = \sum_{k=1}\psi(\frac{x}{k})$$

So that:

$$\ln(x!) - \ln\left(\left\lfloor\frac{x}{2}\right\rfloor!\right)- \ln\left(\left\lfloor\frac{x}{3}\right\rfloor!\right)- \ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right) = $$

$$\psi(x) + \psi\left(\frac{x}{5}\right) - 2\psi\left(\frac{x}{6}\right) + \ldots + \psi\left(\frac{x}{6w-5}\right) + \psi\left(\frac{x}{6w-1}\right) - 2\psi\left(\frac{x}{6w}\right) + \ldots$$

So, for each $w > 1$, it follows that:

$$\psi\left(\frac{x}{6w-5}\right) + \psi\left(\frac{x}{6w-1}\right) \ge 2\psi\left(\frac{x}{6w}\right)$$

So, if there exists $k$ such that:

$$\sum_{w=2}^{k} \psi\left(\frac{x}{6w-5}\right) + \psi\left(\frac{x}{6w-1}\right) - 2\psi\left(\frac{x}{6w}\right) \ge 2\psi\left(\frac{x}{6}\right) - \psi\left(\frac{x}{5}\right)$$

Then it follows that:

$$\ln(x!) - \ln\left(\left\lfloor\frac{x}{2}\right\rfloor!\right)- \ln\left(\left\lfloor\frac{x}{3}\right\rfloor!\right)- \ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right) - \psi(x) \ge 0$$

What will be the problem with this approach? Am I making a mistake in any of my steps?



  • $\begingroup$ Take $x=113$ your inequality at the bottom is not true $\endgroup$ – Ethan May 12 '13 at 1:44
  • $\begingroup$ I am trying to make the argument for $x \ge 2863$. For $x < 2863$, it is often not true. $\endgroup$ – Larry Freeman May 12 '13 at 3:43
  • $\begingroup$ $$\ln(x!)-\ln(x/2!)-\ln(x/3!)-\ln(x/6!)$$ $$=\sum_{n}\psi(\frac{x}{6n-1})+\psi(\frac{x}{6n-3})+\psi(\frac{x}{6n-5})- \psi(\frac{x}{6n})$$ $\endgroup$ – Ethan May 12 '13 at 4:21
  • $\begingroup$ Thanks for writing this. I'm not clear on why it would be $\psi(\frac{x}{6n-3}) - \psi(\frac{x}{6n})$ and not $-2\psi(\frac{x}{6n})$ and where would $\psi(x)$ fit in? I thought it would be $$\psi(x) + \sum\limits_{n}\psi(\frac{x}{6n-1}) + \psi(\frac{x}{6n-5}) - 2\psi(\frac{x}{6n})$$ $\endgroup$ – Larry Freeman May 12 '13 at 5:32
  • 1
    $\begingroup$ What's here looks good to me, although of course the missing step that determines "if there exists $k$" is important. $\endgroup$ – Zander May 12 '13 at 23:48

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