[talanum1 (OP)]

At the "Prime Number Theorem" page, they say in a figure that lim_n->infinity pi(n)/li(n) = 1 and later they say li (n) - pi (n) has infinitely many sign changes.

Which is it.

[evan_au]

Without having calculated it myself, I can see that both can be true.

Approximation (as n→∞)
First, Wikipedia give two approximations to π(n)= "the quantity of prime numbers <n"; one approximation converges much more quickly than the other.
- The poorer approximation π(n) ≈ n/ln(n) always estimates too high (for n>100 or so)
- The better approximation π(n) ≈ ∫1/ln(t) estimates too low (for n=100 to 1,000,000), but is fairly close above that

lim_n→∞ π(n)/∫1/ln(t) = 1

The first statement is a measure of Relative Error.
It says that if you take these two very large numbers (approaching infinity), and divide one by the other, the answer approaches 1. So the Relative Error approaches 0 as n→∞.

∫1/ln(t) - π(n) has infinitely many sign changes

The second statement is a measure of Absolute Error.

Since the density of primes is very low when you get to large n, (the graph shows n up to 10²⁴), prime numbers get very rare. The function π(n) spends most of the time unchanged as n increases. But when it hits another prime number, it will suddenly jump up by 1.

In contrast, ∫1/ln(t) always increases, even when n is not a prime. That means π(n) will sometimes be above it, and sometimes below it. And since there are an infinite number of primes, that "sometimes" is an infinite number of times.

That tells you that ∫1/ln(t) is a pretty good approximation (provided n > 1 million or so).

Perhaps more mind-bending, they also say that the maximum Absolute Error of ∫1/ln(t) grows without limit as n→∞.
- However, this Absolute Error grows much more slowly than π(n), so the Relative Error is infinitesimal.

Here is the link where this is discussed: https://en.wikipedia.org/wiki/Prime_number_theorem#Statement

[talanum1 (OP)]

Can one then say that in the limit of n tending to infinity: pi (n) = li (n).

If yes then it is an easy matter proving the Riemann Hypothesis.

[Bored chemist]

Can one then say that in the limit of n tending to infinity: pi (n) = li (n).

If yes then it is an easy matter proving the Riemann Hypothesis.

One of the small number of bits of the page here
https://en.wikipedia.org/wiki/Prime_number_theorem
which I actually understand says
"So, the prime number theorem can also be written as π(x) ~ Li(x). ".

But I very much doubt that...

If yes then it is an easy matter proving the Riemann Hypothesis.

[evan_au]

...then it is an easy matter proving the Riemann Hypothesis

If you search that Wikipedia page for "Riemann", you will find a lot of hits, mostly around use of the Riemann Zeta function.
- So a lot of mathematicians have studied the Riemann Zeta function in depth
- The Riemann hypothesis is about the Riemann Zeta function
- But it's still not yet proven - so it's not "easy"!

You will also find a few hits where the mathematician has just assumed the Riemann Hypothesis is true in order to provide a better method. Such solutions are provisional, but probably still useful.

[Bored chemist]

I was just judging on the basis that they don't usually offer million dollar prizes for things that are "easy".

[evan_au]

Perhaps more mind-bending, they also say that the maximum Absolute Error of Li(x)=∫1/ln(t) grows without limit as x→∞.

Thinking about this some more... (and viewing the graph)...

When searching through large numbers, primes occur at (pseudo-)random intervals.
- While Li(x) is a long-term average of the expected number of primes.
- So the Absolute Error |Li(x) - π(x)| is actually a (pseudo-)random walk, which can stray large distances from 0
- If it were a true random walk, the standard deviation of absolute error would grow as √x.
- And maximum Absolute Error |Li(x) - π(x)| → ∞ as x → ∞

But this is still compatible with π(x)/Li(x)  → 1 as x → ∞
..since Relative Error (√x)/x → 0 as x → ∞