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Incorrect Wikipedia Article?

  1. Mar 12, 2006 #1
    Hey folks,

    I'm currently studying sequences and the like in Calc 2, and I went to Wikipedia for another explaination about them. The example given in the article located here seems to be incorrect to me.

    The example is this:

    [itex]\sum _{n=0}^{\infty }{2}^{-n} = 2[/itex]

    I was thinking it's equal to zero though, since when n is really large, then the bottom gets really big so the whole fraction would head to zero.

    Am I wrong or is the author wrong?
  2. jcsd
  3. Mar 12, 2006 #2
    This is a series, and not a sequence.
  4. Mar 12, 2006 #3
    Yeah, that's what I meant to say o:) ... but should that change the answer?
  5. Mar 12, 2006 #4


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    Staff Emeritus
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    Gold Member

    Well, yes. A "series", a "sum of a series", a "sequence", and a "limit of a sequence" are all very different things.
  6. Mar 12, 2006 #5
    You're thinking of [itex]\lim_{n \to \infty} 2^{-n}[/itex], which is zero. However [itex]\sum 2^{-n} = 1 + \frac{1}{2} + \frac{1}{4} + \cdots[/itex] isn't zero.
  7. Mar 12, 2006 #6
    Yes, that answer is correct. [tex]1 + \frac{1}{2} + \frac{1}{4} + \cdots[/tex] does indeed equal 2.

    I probably solve simple series like this in a unique way, but I tend to imagine it in the number base 2 (binary). This would essentially be 1.11111111 repeating. This is like our 9.9999 repeating = 10, only that in binary is 2.
  8. Mar 13, 2006 #7


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    Uhmmm, I am sorry to tell you this, but you are wrong, not the author... :tongue2:
    This is a geometric series with the first term 1, and the common ratio r = 1 / 2.
    So apply the formula to find the sum of the first n terms of a geometric series, we have:
    [tex]S_n = a_1 \frac{1 - r ^ n}{1 - r}[/tex]
    Now r = 1 / 2. So |r| < 1, that means:
    [tex]\lim_{n \rightarrow \infty} r ^ n = 0[/tex]
    Now let n increase without bound to get the sum:
    [tex]\sum_{n = 0} ^ {\infty} 2 ^ {-n} = \lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} a_1 \frac{1 - r ^ n}{1 - r} = \frac{a_1}{1 - r} = \frac{1}{1 - \frac{1}{2}} = 2[/tex].
    Can you get this? :)
    @ KingNothing: Have you leant geometric series? We don't need to complicate the problem in binary, though. Just my $0.02.
    Last edited: Mar 13, 2006
  9. Mar 13, 2006 #8
    Doh!! I get it now! Thanks folks... and I apologize for my confusion :(... it's all a little complicated when you first learn it.
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