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1 + 10 + 100 + 1000 + = -1/9

  1. Mar 3, 2010 #1
    [tex]S = 1 + 10 + 100 + 1000 + 10000 + ... [/tex]

    [tex]10S = 10 + 100 + 1000 + 10000 + 100000 + ... [/tex]

    [tex]S - 10S = (1 + 10 + 100 + 1000 + 10000 + ...) - (10 + 100 + 1000 + 10000 + ...) [/tex]

    [tex]-9S = 1 + (10 - 10) + (100 - 100) + (1000 - 1000) + (10000 - 10000) ... [/tex]

    [tex]-9S = 1 + 0 + 0 + 0 + 0 + 0 ...[/tex]

    [tex]-9S = 1 [/tex]

    [tex]S = -1/9[/tex]

    :confused:
    What's wrong (or right) with this?

    Thanks,
    Unit
     
  2. jcsd
  3. Mar 3, 2010 #2

    CRGreathouse

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  4. Mar 3, 2010 #3
  5. Mar 3, 2010 #4

    Hurkyl

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    Also, I believe that sum converges as an ordinary infinite sum in the 2-adics and the 5-adics.

    (And, of course, it does not converge as an ordinary infinite sum in the reals!)
     
  6. Mar 4, 2010 #5

    CRGreathouse

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    I would have [intuitively] expected it to converge in all the p-adics. Am I wrong?
     
  7. Mar 4, 2010 #6

    Hurkyl

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    In any other p-adic field, the terms don't converge to zero!
     
  8. Mar 4, 2010 #7

    CRGreathouse

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    :blushing:
     
  9. Mar 5, 2010 #8

    Char. Limit

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    I'm pretty sure that you can't pair up terms in an infinite sum.
     
  10. Mar 5, 2010 #9

    Redbelly98

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    The way I remember it, proofs like this actually say something like:
    If S exists, then S = 1 + 10 + 100 + ...​
    So if S does not exist, then the remaining statements do not necessarily hold true.
     
  11. Mar 5, 2010 #10
    :smile: Brilliant! I had completely forgotten about variables and their related hypothetical syllogisms. Thanks!
     
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