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Divergence of the sum of the reciprocals of the primes

  1. May 6, 2012 #1
  2. jcsd
  3. May 6, 2012 #2


    They are using...that theorem. If you want to see why is true take into consideration everything's positive

    here and apply logarithms. Of course, you could also google "infinite products"...:>)

    DonAntonio
     
  4. May 6, 2012 #3
    Well, I can say:

    Let [itex]a_n>0, ~n\in \bf{N}[/itex]. We can say [itex]\displaystyle\prod_{n=1}^{+\infty}(1+a_n)[/itex] converge iff converge [itex]\displaystyle\sum_{n=0}^{+\infty}\log(1+a_n)[/itex]. Also, [itex]\displaystyle\sum_{n=0}^{+\infty}\log(1+a_n)[/itex] converge iff converge [itex]\displaystyle\sum_{n=0}^{+\infty}a_n[/itex], because [itex]\displaystyle\lim_{n\to +\infty}\frac{\log(1+a_n)}{a_n} = 1[/itex].

    So, we have: [itex]\displaystyle\prod_{n=1}^{+\infty}(1+a_n)[/itex] converge iff converge [itex]\displaystyle\sum_{n=0}^{+\infty}a_n[/itex].

    BUT, here, I have this minus....?!?!?!?:uhh:
     
  5. May 6, 2012 #4


    Don't worry about that: check theorem 1 here http://tinyurl.com/7cu4k4p , which sends you to check the nice book by

    Knopp "Infinite sequences and series" (see chapter 3, section 7 there)

    DonAntonio
     
  6. May 7, 2012 #5
    OK, it seems well.

    Thanks.
     
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