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Proof question: the sum of the reciprocals of the primes diverges

  1. Jun 13, 2014 #1
    The gist of the approach I took is that∑1/p = log(e^∑1/p) = log(∏e^1/p) and logx→ ∞ as x→∞.
    Proof outline: let ∑1/p = s(x). (...SO I can write this easily on tablet) and note that e^s(x) diverges since e^1/p > 1 for any p and the infinite product where every term exceeds 1 is divergent. Then loge^s(x) diverges as logs as x→∞ would. Thus, since log(e^s(x)= s(x), the sum is found to be divergent
    Last edited: Jun 13, 2014
  2. jcsd
  3. Jun 13, 2014 #2
    That's not right.
  4. Jun 13, 2014 #3

    D H

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    That's obviously incorrect since ##\Pi_{n=1}^{\infty} \exp(2^{-n}) = e##.
  5. Jun 13, 2014 #4
    Wow that isvrry wrong indeed. Thank you
  6. Jun 13, 2014 #5
    Taking the advice i was initally given - by starting with a product represation of the harmonic series has cerainly panned out better. i keptthinking "my method doesn't seem right somewhere..."
  7. Jun 13, 2014 #6


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    Last edited: Jun 13, 2014
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