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Prime number reciprocal sum

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  • Mike
    By no means am I an expert on prime numbers, having only had my first taste of number theory a few semesters ago, but would it be reasonable to conjecture that
    Message 1 of 9 , Sep 19 2:49 PM
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      By no means am I an expert on prime numbers, having only had my first
      taste of number theory a few semesters ago, but would it be
      reasonable to conjecture that the sum of the reciprocals of primes
      (starting at 2 and increasing thereafter) approaches and eventually
      equals a finite number, i.e.

      inf
      (sum) 1/p = r for some r in the real numbers ??
      p=2
      p prime



      Any proof/disproof/thoughts would be appreciated! If someone knew
      the actual number (assuming it exists), that would be most
      fascinating to me.


      Thank you
      Michael Omstead
    • Andy Swallow
      ... Umm, see any introductory book on number theory, and look for Merten s theorem. If you let s(x) be the sum of all reciprocals of primes p
      Message 2 of 9 , Sep 19 2:54 PM
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        On Fri, Sep 19, 2003 at 09:49:39PM -0000, Mike wrote:
        > By no means am I an expert on prime numbers, having only had my first
        > taste of number theory a few semesters ago, but would it be
        > reasonable to conjecture that the sum of the reciprocals of primes
        > (starting at 2 and increasing thereafter) approaches and eventually
        > equals a finite number, i.e.

        Umm, see any introductory book on number theory, and look for Merten's
        theorem. If you let s(x) be the sum of all reciprocals of primes p<=x,
        then

        s(x) = loglog x + C + o(1),

        where C is some constant, and o(1) denotes a term which approaches zero
        as x-->oo. So no, it doesn't converge. It diverges very slowly.

        Andy
      • Jose Ramón Brox
        In fact, the easiest way I know to prove that the prime numbers are an infinity is that the sum of their reciprocals is divergent. But you know, if you sum the
        Message 3 of 9 , Sep 19 3:06 PM
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          In fact, the easiest way I know to prove that the prime numbers are an infinity is that the sum of their reciprocals is divergent.

          But you know, if you sum the reciprocals of the twin primes, then the sum is convergent! And the real number that is the limit is called Brun's constant (go to the Chris Caldwell's glossary to know more).

          Jose Brox

          ----- Original Message -----
          From: Andy Swallow
          To: primenumbers@yahoogroups.com
          Sent: Friday, September 19, 2003 11:54 PM
          Subject: Re: [PrimeNumbers] Prime number reciprocal sum


          On Fri, Sep 19, 2003 at 09:49:39PM -0000, Mike wrote:
          > By no means am I an expert on prime numbers, having only had my first
          > taste of number theory a few semesters ago, but would it be
          > reasonable to conjecture that the sum of the reciprocals of primes
          > (starting at 2 and increasing thereafter) approaches and eventually
          > equals a finite number, i.e.

          Umm, see any introductory book on number theory, and look for Merten's
          theorem. If you let s(x) be the sum of all reciprocals of primes p<=x,
          then

          s(x) = loglog x + C + o(1),

          where C is some constant, and o(1) denotes a term which approaches zero
          as x-->oo. So no, it doesn't converge. It diverges very slowly.

          Andy


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        • Andy Swallow
          ... Call me old fashioned, but surely Euclid s argument is a tad better and easier for proving the infinitude of the primes... Andy
          Message 4 of 9 , Sep 19 3:13 PM
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            On Sat, Sep 20, 2003 at 12:06:18AM +0200, Jose Ram?n Brox wrote:
            > In fact, the easiest way I know to prove that the prime numbers are an
            > infinity is that the sum of their reciprocals is divergent.

            Call me old fashioned, but surely Euclid's argument is a tad better and
            easier for proving the infinitude of the primes...

            Andy
          • Paul Leyland
            ... Ok: you are old fashioned. I like this proof of the infinitude of primes. First, observe that there are an infinite number of distinct integers of the form
            Message 5 of 9 , Sep 23 4:07 AM
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              > From: Andy Swallow [mailto:umistphd2003@...]
              > Sent: 19 September 2003 23:13
              > On Sat, Sep 20, 2003 at 12:06:18AM +0200, Jose Ram?n Brox wrote:
              > > In fact, the easiest way I know to prove that the prime
              > numbers are an
              > > infinity is that the sum of their reciprocals is divergent.
              >
              > Call me old fashioned, but surely Euclid's argument is a tad
              > better and
              > easier for proving the infinitude of the primes...

              Ok: you are old fashioned.


              I like this proof of the infinitude of primes.

              First, observe that there are an infinite number of distinct integers
              of the form F_n = 2^2^n+1 where n is an integer >= 0.

              Next, observe that F_{n+1} - 2 = (2^2^{n+1} - 1) and that this
              expression forms the difference between two squares. We know how to
              factor such expressions.

              In particular, F_{n+1} - 2 = (2^2^n + 1) * (2^2^n - 1).

              But the terms in the product are just F_n and F_n - 2.

              So, F_{n+1} is coprime to F_n because the latter divides F_{n+1} - 2.

              We've now demonstrated an infinite series of integers, none of which
              share a prime factor, so there must be an infinite number of primes.


              Take a bow, M. Pierre Fermat.

              Paul
            • mikeoakes2@aol.com
              In a message dated 23/09/03 12:09:28 GMT Daylight Time, ... Have you? F_{n+1} is coprime to F_n, and F_{n+2} is coprime to F_{n+1), but is F_{n+2} necessarily
              Message 6 of 9 , Sep 23 4:29 AM
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                In a message dated 23/09/03 12:09:28 GMT Daylight Time,
                pleyland@... writes:

                >So, F_{n+1} is coprime to F_n because the latter divides F_{n+1} - 2.
                >
                >We've now demonstrated an infinite series of integers, none of which
                >share a prime factor, so there must be an infinite number of primes.

                Have you?
                F_{n+1} is coprime to F_n, and F_{n+2} is coprime to F_{n+1), but is F_{n+2}
                necessarily coprime to F_n?

                Mike



                [Non-text portions of this message have been removed]
              • Andy Swallow
                ... Rather amusing really. But of course, the proof that Paul presented can be more properly written out to show that F_n and F_{n+k} have no common factors
                Message 7 of 9 , Sep 23 4:58 AM
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                  > >So, F_{n+1} is coprime to F_n because the latter divides F_{n+1} - 2.
                  > >
                  > >We've now demonstrated an infinite series of integers, none of which
                  > >share a prime factor, so there must be an infinite number of primes.
                  >
                  > Have you?
                  > F_{n+1} is coprime to F_n, and F_{n+2} is coprime to F_{n+1), but is F_{n+2}
                  > necessarily coprime to F_n?

                  Rather amusing really. But of course, the proof that Paul presented can
                  be more properly written out to show that F_n and F_{n+k} have no common
                  factors for any k>0.

                  So that a proof using Fermat numbers. Both that method and Euclid's can
                  be used to estimate the number of primes up to a certain point. And both
                  estimates are terrible when compared to more 'modern' methods. But just
                  for the record, the estimate given by Euclid's argument is a lot closer
                  to the truth than that given by considering Fermat numbers...

                  Andy
                • Paul Leyland
                  Yes it is. Think about it. To get you started, try factoring one of the terms in the product I gave you. Paul ________________________________ From:
                  Message 8 of 9 , Sep 23 4:59 AM
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                    Yes it is. Think about it.

                    To get you started, try factoring one of the terms in the product I gave you.

                    Paul



                    ________________________________

                    From: Mikeoakes2@... [mailto:Mikeoakes2@...]
                    Sent: 23 September 2003 12:29
                    To: Paul Leyland
                    Cc: primenumbers@yahoogroups.com
                    Subject: Re: [PrimeNumbers] Prime number reciprocal sum


                    In a message dated 23/09/03 12:09:28 GMT Daylight Time, pleyland@... writes:

                    >So, F_{n+1} is coprime to F_n because the latter divides F_{n+1} - 2.
                    >
                    >We've now demonstrated an infinite series of integers, none of which
                    >share a prime factor, so there must be an infinite number of primes.

                    Have you?
                    F_{n+1} is coprime to F_n, and F_{n+2} is coprime to F_{n+1), but is F_{n+2} necessarily coprime to F_n?

                    Mike




                    [Non-text portions of this message have been removed]
                  • navid.altaf@gstt.sthames.nhs.uk
                    Just wondering. Is the following argument right? Consider a continuous list of primes starting 2 ie 2,3,5,7..............p 1. we can define composites and
                    Message 9 of 9 , Sep 23 6:59 AM
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                      Just wondering. Is the following argument right?

                      Consider a continuous list of primes starting 2 ie 2,3,5,7..............p

                      1. we can define composites and primes in the range >p <p^2

                      2. there will always be at least one prime in the range >p <p^2.
                      Therefore, there must be an infinitude of prime numbers.




                      Navid



                      -----Original Message-----
                      From: Paul Leyland [mailto:pleyland@...]
                      Sent: 23 September 2003 12:08
                      To: primenumbers@yahoogroups.com; Andy Swallow
                      Subject: RE: [PrimeNumbers] Prime number reciprocal sum



                      > From: Andy Swallow [mailto:umistphd2003@...]
                      > Sent: 19 September 2003 23:13
                      > On Sat, Sep 20, 2003 at 12:06:18AM +0200, Jose Ram?n Brox wrote:
                      > > In fact, the easiest way I know to prove that the prime
                      > numbers are an
                      > > infinity is that the sum of their reciprocals is divergent.
                      >
                      > Call me old fashioned, but surely Euclid's argument is a tad
                      > better and
                      > easier for proving the infinitude of the primes...

                      Ok: you are old fashioned.


                      I like this proof of the infinitude of primes.

                      First, observe that there are an infinite number of distinct integers
                      of the form F_n = 2^2^n+1 where n is an integer >= 0.

                      Next, observe that F_{n+1} - 2 = (2^2^{n+1} - 1) and that this
                      expression forms the difference between two squares. We know how to
                      factor such expressions.

                      In particular, F_{n+1} - 2 = (2^2^n + 1) * (2^2^n - 1).

                      But the terms in the product are just F_n and F_n - 2.

                      So, F_{n+1} is coprime to F_n because the latter divides F_{n+1} - 2.

                      We've now demonstrated an infinite series of integers, none of which
                      share a prime factor, so there must be an infinite number of primes.


                      Take a bow, M. Pierre Fermat.

                      Paul


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