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Re: Help with Limit of a zeta function

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  • Andrey Kulsha
    Hello Don, ... this sum is convergent, but I can t invent how to calculate it. As for error term, I think, if you take first k terms of the sum instead of
    Message 1 of 2 , Jun 3, 2003
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      Hello Don,

      > Can someone tell me if the following function is convergent and if so
      > what it is convergent to with some degree of precision and an error
      > term?
      >
      > For {n=[1->oo], P=primes[2->oo]} sum{n/Pn^2}

      this sum is convergent, but I can't invent how to calculate it. As for error term, I think, if you take first k terms of the sum instead of infinity, the error will be about 1/log(k).

      Best wishes,

      Andrey


      [Non-text portions of this message have been removed]
    • Don Christmann
      Can someone tell me if the following function is convergent and if so what it is convergent to with some degree of precision and an error term? For {n=[1- oo],
      Message 2 of 2 , Jun 3, 2003
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        Can someone tell me if the following function is convergent and if so
        what it is convergent to with some degree of precision and an error
        term?

        For {n=[1->oo], P=primes[2->oo]} sum{n/Pn^2}

        It is bascially the Prime zeta P2 but for each next p(rime) the
        number on top also increments ie. -

        1/2^2 + 2/3^2 + 3/5^2 + 4/7^2 + 5/11^2 + 6/13^2 + 7/17^2 + ...

        Also, is this a function already known by some name? Perhaps in
        another form - double zeta-P2, second-order zeta-P2, something?

        Thanks.
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