Is the Example for Series in Wikipedia Incorrect?

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Discussion Overview

The discussion centers around the example of a series presented in a Wikipedia article, specifically the infinite series \(\sum _{n=0}^{\infty }{2}^{-n}\). Participants explore whether the example is correct, with a focus on understanding the distinction between sequences and series in calculus.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • One participant questions the correctness of the Wikipedia example, believing the series converges to zero as \(n\) increases.
  • Another participant clarifies that the example refers to a series, not a sequence, and emphasizes the difference between these concepts.
  • A participant explains that the limit of the terms \(2^{-n}\) approaches zero, but the sum of the series converges to 2, providing the series' expansion as \(1 + \frac{1}{2} + \frac{1}{4} + \cdots\).
  • Further elaboration on the geometric series formula is provided, detailing how to compute the sum and confirming the result of 2.
  • One participant reflects on their unique approach to understanding the series using binary representation, relating it to repeating decimals.
  • A later reply acknowledges the initial confusion and expresses gratitude for the clarification, indicating a learning moment.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial claim regarding the correctness of the Wikipedia example, but there is agreement on the correct interpretation of the series and its sum. The discussion reveals differing levels of understanding regarding series and sequences.

Contextual Notes

Some participants express confusion regarding the concepts of series and sequences, indicating a potential gap in foundational knowledge. The discussion includes various interpretations of the series and its convergence properties.

Weather Freak
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Hey folks,

I'm currently studying sequences and the like in Calc 2, and I went to Wikipedia for another explanation about them. The example given in the article http://en.wikipedia.org/wiki/Series_%28mathematics%29" 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?
 
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This is a series, and not a sequence.
 
Yeah, that's what I meant to say o:) ... but should that change the answer?
 
Well, yes. A "series", a "sum of a series", a "sequence", and a "limit of a sequence" are all very different things.
 
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.
 
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.
 
Weather Freak said:
Am I wrong or is the author wrong?
Uhmmm, I am sorry to tell you this, but you are wrong, not the author... :-p
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:
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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