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Is the sum of all primes = 13?

  1. Jul 24, 2011 #1
    I posted this to Dr. Math but I'm too excited to wait for their response.

    OK, so start with the following equation, http://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_…#Summability" by Ramanujan and Euler:

    1 + 2 + 3 + 4 + ... = -1/12

    Weird, yes, but there are http://planetmath.org/encyclopedia/PAdicValuation.html" [Broken] under which these kinds of expressions are meaningful.

    OK, so let's call the series s:

    s = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + ...

    Now, we can gather all multiples of 2 into one term:

    s = 1 + 2s + 3 + 5 + 7 + 9 + 11 + ...

    Similarly, we can gather all multiples of 3:

    s = 1 + 2s + 3s + 5 + 7 + 11 + ...

    Continue sieving this way in a manner similar to Eratosthenes until you have:

    s = 1 + s * ( sum{all primes p} p )

    Rearranging:

    (s - 1) / s = sum{all primes p} p

    But since we already know the value of s = -1/12:

    (-13/12) / (-1/12) = sum{all primes p} p

    sum{all primes p} p = 13

    QED

    I see no flaw in my reasoning.

    Clayton
     
    Last edited by a moderator: May 5, 2017
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  3. Jul 24, 2011 #2

    D H

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    You are adding numbers such as 6 and 15 twice, numbers such as 30 and 1001 three times, etc.
     
  4. Jul 24, 2011 #3
    Crap. Oh well, so much for that.
     
  5. Jul 24, 2011 #4

    Hurkyl

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    p-Adic valuations don't help with a series like this. You need some sort of alternate definition of summation.

    One rather typical way to give meaning to the sum of all primes is to consider the function

    [tex]f(s) = \sum_p p^s[/tex]​

    The relevant facts I expect to be true are that this sum is well-defined and analytic for all complex s with Re(s) < -1.

    Then, you use analytic continuation to extend it to as much of the complex plane as possible. With luck, it might even be a single-valued function!

    (If you sum over all positive integers instead of just the primes, you get, more or less, the Riemann Zeta function)
     
  6. Jul 24, 2011 #5
    How come this doesn't diverge, even though s is complex? If Im(s) = 0, then it seems like it must be divergent.
     
  7. Jul 24, 2011 #6

    Mute

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    It does diverge. The series requires Re(s) < -1 (for the Riemann zeta function, at least). For Re(s) > 1, analytic continuation is required, and the sum does not represent the function for the analytically continued region (although sometimes people speak as though the sum evaluates to the value given by the analytic continuation of the function).
     
  8. Jul 24, 2011 #7
    Ah, I get it, when Re(s) < -1, it satisfies the criterion for convergence of the Riemann zeta function - yes, this is not the zeta function but I can see it's closely related to it - so there's probably a way to prove that "If Riemann zeta function converges at s then Sum p^s converges at -s". I'm pretty shaky on analytic continuation, I get the basic idea but don't understand the mechanics of it.

    Clayton -
     
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