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Re: LI(Pn#)=P(n-1)# (approximately)

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  • Werner D. Sand
    Proof: You don t need Li, PNT is enough. We start with a form of PNT: sum(ln p)(p lim(n- inf)(sum(ln p)(p lim(n- inf)(pn /
    Message 1 of 2 , Mar 16, 2007
      Proof:

      You don't need Li, PNT is enough. We start with a form of PNT:
      sum(ln p)(p<=pn) ~ pn. ==>
      lim(n->inf)(sum(ln p)(p<=pn) / pn) = 1 ==>
      lim(n->inf)(pn / sum(ln(p)(p<=pn) = 1 ==>
      lim(n->inf)(pn / ln(pn#)(p<=pn) = 1 ==>
      lim(n->inf)(pn*p(n-1)#) / (ln(pn#)*p(n-1)#) = 1 ==>
      lim(n->inf)(pn#) / (ln(pn#)*p(n-1)#) ) = 1 ==>
      lim(n->inf)(pn#) / ln(pn#) = p(n-1)# = PNT, qed.

      Werner
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