Pi[x] >=loglogx

  • #1
I am going through Hardy's book on number theory.The following theorem I do not understand.

theorem 10: pi[x] >= loglog x
where pi[x] is the prime counting function
and >= stands for greater than or equal to

The arguments written in the book are very compact.please help .
 

Answers and Replies

  • #2
shmoe
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Do you follow any of it?

Do you understand how they derived [tex]p_{n}<2^{2^n}[/tex] ?

this is an important step. The rest just follows from pi(x) being increasing, and also [tex]\pi(p_n)=n[/tex] which they use but don't explicitly mention.
 
  • #3
CRGreathouse
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Bertrand's postulate?
 
  • #4
shmoe
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Nope! the bound of p_n above is much weaker than Bertrand's will give you. It's correspondingly simpler to prove though, it follows from a slight adaptation of Euclid's proof there are infinitely many primes (in case anyone who hasn't seen it wants to give it a stab)
 

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