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It's well know that $a \frac{x}{\log x}\le \pi(x) \le b \frac{x}{\log x}$ for some constant $a,b \in \mathbb{R}$.

There are a lot of ideas to prove it, but I've saw an exercise with following statement :

Prove that if $\sum_{p \le x} \frac{\log p}{p} - \log x = \lambda + o(1)$ , then asymptotic law holds.

I don't know where is the connection between them?

openspace
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  • Related: https://math.stackexchange.com/a/3374521/44121 – Jack D'Aurizio Sep 30 '19 at 18:33
  • @JackD'Aurizio I thought there is inequality, not asymptotic behavior. – openspace Sep 30 '19 at 18:46
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    There's not much difference: once you prove that $\pi(x)\frac{\log x}{x}\to 1$ by the given asymptotics, it follows that for any $x$ large enough the inequality $(1-\varepsilon)\leq \pi(x)\frac{\log x}{x}\leq(1+\varepsilon)$ holds. – Jack D'Aurizio Sep 30 '19 at 18:49

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If we know that $$ \sum_{p\leq x}\frac{\log p}{p} = \log(x)-\lambda+o(1) $$ holds, we also know that $$ \sum_{x/2<p\leq x}\frac{\log p}{p} = \log(2)+o(1) $$ holds. All the terms of the above sum are bounded between $\frac{\log(x/2)}{x/2}$ and $\frac{\log(x)}{x}$, so $$ \log(2)+o(1) \geq \left[\pi(x)-\pi(x/2)\right]\frac{\log x}{x}, $$ $$ \log(2)+o(1) \leq \left[\pi(x)-\pi(x/2)\right]\frac{\log(x)}{x/2} $$ or $$ (\log(2)+o(1))\frac{x}{2\log(x)}\leq\pi(x)-\pi(x/2) \leq (\log(2)+o(1))\frac{x}{\log x}.$$ By replacing $x$ with $x/2,x/4,x/8,\ldots$ and summing all these inequality we get $$ A\frac{x}{\log x}\leq \pi(x) \leq B\frac{x}{\log x} $$ as wanted, with $A\approx \log(2)$ and $B\approx 2\log(2)$.

Jack D'Aurizio
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