- #1
peteryellow
- 47
- 0
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.
$P(x) = \sum_{k=1}^{\infty} \frac 1k \pi(x^{1/k})$
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.