# Pi[x] >=loglogx

by AlbertEinstein
Tags: >loglogx
 Sci Advisor HW Helper P: 1,995 Do you follow any of it? Do you understand how they derived $$p_{n}<2^{2^n}$$ ? this is an important step. The rest just follows from pi(x) being increasing, and also $$\pi(p_n)=n$$ which they use but don't explicitly mention.