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The Sum to Infinety of GP's

  1. Sep 7, 2004 #1
    In a geometric progression you can find the sum to infinety is some series, for example 4,2,1,... where the common factor is 1/2. The sum to infinety will then be, 8, it says in my book, but I can only think of it as very, very close to 8, not eight exactly. How is it? Is the sum to infinety 8 or just very close to eight?
  2. jcsd
  3. Sep 7, 2004 #2
    Sum of a GP = a(1-r^n)/(1-r) where a is the first number in the sequence, r is the common ratio and n is the term number.

    When -1<r<1 and n approaches infinity r^n approaches 0. Therefore Sum to infinity = a/(1-r)

    Therefore Sum to infinity = 4/(1-0.5) = 8
  4. Sep 7, 2004 #3

    matt grime

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    The sum *is* eight. It is the limit of the finite subsums. If it weren't eight but were less than 8, then yo'ud have a problem since one of the (increasing) finite subsums would be greater than your preferred infinite sum. It is a property of the real number system that the sum is 8. It is, by definition, 8 there is no contention about that, if you think it is something different then you don't understand what the words mean.
  5. Sep 7, 2004 #4


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    If you were to stop somewhere short of "infinity", say summing up to n= 10000000, then the answer, one of the "subsums" that matt grime referred to (I would say "partial sum") would be slightly less than 8. Summing all terms, that is, never stopping, will give exactly 8.
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