Recall Gauss' estimate for \(\pi(x)\), the logarithmic integral function.

It wasn't too bad. But we were hoping we could get a little closer. So, among several other things, we tried \[Li(x)-\frac{1}{2}Li(\sqrt{x})\; .\] And this was indeed better.

So one might think one could keep adding and subtracting \[\frac{1}{n}Li(x^{1/n})\] to get even closer, with this start to the pattern.

As it turns out, that is not quite the right pattern. In fact, the minus sign comes from \(\mu(2)\), not from \((-1)^{2+1}\), as usually is the case in series like this!

However, it should be just as plain that this approximation doesn't really add a lot beyond \(k=3\). In fact, at \(x=1000000\), just going through \(k=3\) gets you within one of \(\sum_{j=1}^\infty\frac{\mu(j)}{j}Li(x^{1/j})\). So this is not enough to get a computable, exact formula for \(\pi(x)\).

Questions this might raise:

• So where does this Moebius \(\mu\) come from anyway?
• What else is there to the error \[\left|\pi(x)-Li(x)\right|\] anyway?
• What does this have to do with winning a million dollars?
• Are there connections with things other the just \(\pi(x)\)?