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Re: 2/p +..

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  • gulland68
    I realized my error almost the moment I sent the message - but thanks to those who replied. Tom ... times ... 3, can ... 2/5 + 2/7 + ... the ... then sum. I
    Message 1 of 7 , Apr 21, 2004
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      I realized my error almost the moment I sent the message - but thanks
      to those who replied.
      Tom

      --- In primenumbers@yahoogroups.com, mikeoakes2@a... wrote:
      > In a message dated 20/04/04 18:50:06 GMT Daylight Time,
      > benbradley@m... writes:
      >
      >
      > > >> If you take every prime in sequence and make a sequence of 2
      times
      > > >> the reciprocal of each, then if you exclude the primes 2 and
      3, can
      > > >> it be shown that the sum of all of them will never total 1?
      > > >>
      > > >> thanks,
      > > >> Tom
      > > >
      > > >I'm not sure I understand your question. Do you want to compute
      2/5 + 2/7 +
      > > >2/11 + 2/13 + ...? In that case, the sum of these four terms of
      the
      > > sequence
      > > >already exceeds 1.
      > >
      > > Apparently he means multiply by 2, THEN take the reciprocal,
      then sum. I
      > > wrote a program to do this, and I found the sum exceeds 1 at this
      prime:
      > >
      > > 483281 1.000000724689930
      > >
      >
      > Here's a proper proof, involving no computation at all.
      >
      > Suppose that, for some prime P >= 5,
      > 1/(2*5) + 1/(2*7) + 1/(2*11) + .. + 1/(2*P) = 1.
      > Multiply both sides by the product of all primes < P.
      > Eevery term except the one before the "=" sign is an integer.
      > Contradiction.
      > Q.E.D.
      >
      > -Mike Oakes
      >
      >
      >
      > [Non-text portions of this message have been removed]
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