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Math Help - Why is the sum of 1-2+3-4+... is 1/4?

  1. #16
    Member mohammadfawaz's Avatar
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    Clearly this diverges by nth term test. But what i'm not understanding is that the proofs provided on wikipedia make sense also! they even provide a proof to the following: 1-1+1-1+1-1+... = \frac{1}{2}!!! HOW IN THE WORLD CAN THIS BE EXPLAINED?
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  2. #17
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    Quote Originally Posted by maddas View Post
    (Also, Wilmer, you have wrong series; the signs alternate every term, not every two terms. The partial sums are actually 0,1,-1,2,-2,3,-3,...)
    NO I don't: I'm crearing my own!
    Will be known as The Wilmer Paradox.
    I'll become famous, and will be buried next to Jethro Clampett...
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  3. #18
    Super Member Bacterius's Avatar
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    Quote Originally Posted by chengbin View Post
    Read this

    1 ? 2 + 3 ? 4 + - Wikipedia, the free encyclopedia

    I don't know enough math to understand the "proofs" in that article. Can anyone explain the number manipulation method? It seems like the proof that requires the least amount of math knowledge to understand.
    What ? Now I must tell you that I am not convinced at all. Your expression is equivalent to :

    \sum_{n = 1}^{\infty} -n(-1)^n

    And this sum diverges. So how can it be equal to \frac{1}{4} ? Unless there is some secret mathematical trick that I am not aware of, I don't understand what's up with this statement which is by any means wrong.

    Any clues ?
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  4. #19
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    @Wilmer

    lol

    Well, without any semblance of rigour, write S:=-1-1+1+1-1-1+\cdots. It satisfies -(S+1+1) = S so S=1. Let T:=-2-3+4+5-6-7+\cdots. Adding these term by term gives T+S = -3-4+5+6-7-8+\cdots. Adding S again gives T+2S = -4-5+6+7-8-9+\cdots = -2-3-T. Therefore T=-7/2, if I have made no error. Therefore the Wilmer paradox is 1-2-3+4+5-\cdots = 1-\frac 72 = -\frac 52 :]
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  5. #20
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    Quote Originally Posted by Bacterius View Post
    Any clues ?
    Yes, read the wikipedia article.
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  6. #21
    Super Member Bacterius's Avatar
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    Quote Originally Posted by mohammadfawaz View Post
    Clearly this diverges by nth term test. But what i'm not understanding is that the proofs provided on wikipedia make sense also! they even provide a proof to the following: 1-1+1-1+1-1+... = \frac{1}{2}!!! HOW IN THE WORLD CAN THIS BE EXPLAINED?
    Okay let's do this one quickly.

    1 - 1 + 1 - 1 + 1 - \cdots = (1 - 1) + (1 - 1) + 1 - \cdots = 0 + 0 + 1 - \cdots \color{red}{\neq \frac{1}{2}}

    Well, that's what I think. I don't care what all those integrals on Wikipedia tell me, for me this sequence is definitely equal to 1 for odd n and 0 for even n, regardless of the value of n. Even if n goes towards infinity.

    No one can provide a proof for something that makes no sense. I don't have such a high level of abstraction, sorry. I'm just going to leave this one unsolved because this is getting really confusing. It's a paradox after all. And I don't seem to be prepared to embrace the fact that standard mathematics are not enough to answer this.
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  7. #22
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    We are not talking about the partial sums. See Divergent series - Wikipedia, the free encyclopedia
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  8. #23
    Super Member Bacterius's Avatar
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    Quote Originally Posted by maddas View Post
    We are not talking about the partial sums. See Divergent series - Wikipedia, the free encyclopedia
    Okay maddas, now I'm going to ask you something.
    The OP's sum is :
    - divergent under standard tests.
    - convergent using other methods.

    How can a sum possibly be divergent and convergent ? There must be one of the methods used that fails. Unless I fail. But I'm going to give up because I feel I suck at this kind of mathematics. I'm not even able to understand this stuff even after reading twice the Wikipedia article
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  9. #24
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    It depends on what space you are talking about. Another good example is
    \lim_{n \to \infty} \sin nx
    This has has no limit in the normal sense, but in the sense of distributions the limit is zero.
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  10. #25
    Super Member Bacterius's Avatar
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    Quote Originally Posted by lvleph View Post
    It depends on what space you are talking about. Another good example is
    \lim_{n \to \infty} \sin nx
    This has has no limit in the normal sense, but in the sense of distributions the limit is zero.
    What is x ? Is it a constant ?
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  11. #26
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    x\in \mathbb{R} \backslash\{0\}
    Last edited by lvleph; April 25th 2010 at 11:52 AM.
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  12. #27
    Member mohammadfawaz's Avatar
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    So the definition of the sum of a series as the limit of the partial sum is not actually correct? Or not always correct? I that sense, what we know about summation of series and all the tests we use are based on that definition. For the given series to converge, there must be another definition to use!
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  13. #28
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    A definition cannot be correct or incorrect. It can only be useful, or natural, or motivated, or ...

    This thread really went off the deep end at some point.
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  14. #29
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    Quote Originally Posted by maddas View Post
    This thread really went off the deep end at some point.
    ...at the beginning
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  15. #30
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    It's my fault.
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