Prime Number Theorem: Proving Equivalence of $P(x)$ and $Li(x)$

[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.

 

[CRGreathouse]

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

 

[peteryellow]

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

 

[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 ~)

 

[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.

 

[CRGreathouse]

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

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)).

 

[peteryellow]

yes, but how.