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Firoozbakht QED?

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  • John W. Nicholson
    The Firoozbakht s conjecture (1982) is equal to: (p_(n+1))^(n)
    Message 1 of 2 , Apr 8, 2014

      The Firoozbakht's conjecture (1982) is equal to:

      (p_(n+1))^(n) < (p_n)^(n+1).

      Then the natual log is:

      n*ln(p_(n+1)) < (n+1)*ln(p_n).

      Now,
      ln(p_n) <= ln(n) + ln(ln(n)) + 1, for n >= 2. (Dusart 2010)

      And, because p_n >= n*ln(n), for n >= 2;  
      (Dusart 1999)
      the nat log of p_(n+1) is: 
      ln(n+1) + ln(ln(n+1)) <= ln(p_(n+1)), for n >= 2.

      So,
      n*(ln(n+1) + ln(ln(n+1))) < (n+1)*(ln(n) + ln(ln(n)) + 1).
      Divided by n*(ln(n) + ln(ln(n)) + 1):
      (ln(n+1) + ln(ln(n+1)))/(ln(n) + ln(ln(n)) + 1) < (n+1)/n

      This inequality is true because the left-side increases slower than the right-side.

      Is this a QED?
       
      John W. Nicholson
    • djbroadhurst
      ... No. You failed to understand the concept of inequality. Let A(n)=(n+1)/n; B(n)=log(prime(n+1))/log(prime(n));
      Message 2 of 2 , Apr 10, 2014
        "reddwarf2956" asked:

        > Is this a QED?

        No. You failed to understand the concept of inequality. Let

        A(n)=(n+1)/n;
        B(n)=log(prime(n+1))/log(prime(n));
        C(n)=(log(n+1)+log(log(n+1)))/(log(n)+log(log(n))+1);

        For n > 1, it is easy to prove that
        A(n) > C(n) and
        B(n) > C(n).

        At elementary school you should have learnt that those two
        inequalities do not determine which of A(n) and B(n) is the
        larger.

        David
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