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Re: Real Count-Down Prime

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  • Andrey Kulsha
    Hello! ... Sorry, there was a typo: k goes from 2 to infinity. ... It converges for any n 1. You may calculate an integral(f(k)*dk,C,+infinity) to check the
    Message 1 of 8 , Jul 3, 2001
      Hello!

      Andrey Kulsha wrote:

      > The series
      >
      > sum(1/(klogk),k,1,+infinity)
      >
      > diverges.

      Sorry, there was a typo: k goes from 2 to infinity.

      Phil Carmody wrote:

      > OK, that does seem to make sense. Thanks.
      > (Am I right in thinking that the sum of 1/(k.log(k)^n)
      > would diverge for any constant n?)

      It converges for any n>1.

      You may calculate an integral(f(k)*dk,C,+infinity) to check
      the series

      sum(f(k),k,C,+infinity),

      if f(k) is smooth for C<=k<+infinity.

      Best wishes,

      Andrey
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