Prime number theorem

1. Nov 9, 2008

peteryellow

We have that

$P(x) = \sum_{k=1}^{\infty} \frac 1k \pi(x^{1/k})$
and
$Li(x) = \int_2^n \frac {dt}{\log t}$

And the prime number theorem is:

$$\pi(n) \sim \frac{n}{\log n }$$
I want to show that $$P(x) \sim Li(x)$$ is equivalent to prime number theorem.

Can some body please help me with this.

2. Nov 9, 2008

CRGreathouse

\sim is transitive, so it suffices to show that
$$\operatorname{Li}(x)\sim x/\log(x)$$

3. Nov 9, 2008

peteryellow

I have shown this but still I need to show that P(x) \sim Li(x).

4. Nov 9, 2008

CRGreathouse

Then you're done:
1. $$\pi(x)\sim x/\log(x)$$ (Prime Number Theorem)
2. $$\operatorname{Li}(x)\sim x/\log(x)$$ (you said you proved it already)
3. $$\pi(x)\sim\operatorname{Li}(x)$$ (by transitivity of ~)

5. Nov 10, 2008

peteryellow

I wish I was done but I am not how can I prove that P(x) \sim Li(x).

P is different from /pi. Give any suggestion how can I prove this.

6. Nov 10, 2008

CRGreathouse

Ah, sorry, I forgot your notation. But P(x) is just pi(x) plus some insignificant terms. It suffices to show that P(x) = pi(x) + O(sqrt(x)).

7. Nov 10, 2008

peteryellow

yes, but how.

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