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PhoenixFire

  • 3 years ago

I saw this a little while ago and wanted to figure out how to prove that it's invalid. \[1+2+4+8+...+\infty= -1\] Proof is as follows: Knowing that \[1 * x=x\] and that \[2-1=1\] we can say that \[(2-1)(1+2+4+8+...+\infty)= -1\]expanding the brackets you get\[2+4+8+16+...+\infty- 1 - 2-4-8-...-\infty= -1\] everything from 2 up cancels leaving \[-1=-1\]

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  1. PhoenixFire
    • 3 years ago
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    However, if you do it for anything up to but NOT including infinity it is wrong. \[(2-1)(1+2+4)=2+4+8-1-2-4=8-1=7\]Clearly not -1.

  2. PhoenixFire
    • 3 years ago
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    Something to do with Divergent Series. but I don't understand it. \[\sum_{n=0}^{\infty}2^n\] So I'm basically looking for a way to disprove the above claim for infinity.

  3. UnkleRhaukus
    • 3 years ago
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    \[\infty-\infty\neq0\]

  4. PhoenixFire
    • 3 years ago
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    What do you get in the case of \[\infty - \infty=?\]Is there some law or rule that you can refer me to that explains that it's not equal to zero?

  5. UnkleRhaukus
    • 3 years ago
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    http://www.vitutor.com/calculus/limits/indeterminate_forms.html

  6. UnkleRhaukus
    • 3 years ago
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    infinity is not a number so you cant always treat it like a number, some times you indeterminate forms

  7. PhoenixFire
    • 3 years ago
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    Well that makes sense now. I've always treated infinity as something unique and not a number, but when it came to this my brain got fried. Thanks, @UnkleRhaukus

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