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This Proof is right about sums and limits?

  1. Dec 3, 2015 #1
    hello, sorry for bad English, i have a question.
    if we consider the following equations and we take natural values note that tend 2
    x-1=0 -----------------> x = 1
    x^2-x-1=0 -----------------> x = 1.618033988 (golden ratio)
    x^3-x^2-x-1=0 -----------------> x = 1.839286755
    x^4-x^3-x^2-x-1=0 -----------------> x = 1.927561975
    x^5-x^4-x^3-x^2-x-1=0 -----------------> x = 1.965948236
    x^6-x^5-x^4-x^3-x^2-x-1=0 -----------------> x = 1.983582843

    then we can assume that when the equation has infinite terms the answer is 2.
    and reaches the following "proof" and let me know if it's right:
    1 image:https://gyazo.com/49a46e56fb19b4ec7aa21594e4e78cd1 [Broken]
    2 image:https://gyazo.com/c6b99485e6d0271c1c0bbdbaaca29d54 [Broken]

    Besides knowing if this is OK too I wonder if this is what is called "inductive method"

    thanks.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Dec 3, 2015 #2

    micromass

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    No, you have a divergent series.
     
  4. Dec 3, 2015 #3
    I don't like your proof! If you have
    x^n = 1 + x+ x^2 +...+x^(n-1)
    using the formula for the sum of a GP you get
    x^n = (1-x^n)/(1-x), and so
    x - 2 + 1/x^n =0
    now you can see that as n tends to infinity x gets as close as you like to 2.
     
  5. Dec 4, 2015 #4
    I kept working and I think it gets to the real proof
    https://gyazo.com/aeee169696eebe6c4357520f9dbaa837 [Broken]
     
    Last edited by a moderator: May 7, 2017
  6. Dec 4, 2015 #5

    micromass

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    You cannot use values like ##\infty## or ##\infty-2## like this. Ever.
     
  7. Dec 4, 2015 #6
    I have a question.
    How did you get in here
    x^n = (1-x^n)/(1-x)
    to here
    x - 2 + 1/x^n =0
    my poor mind can not compute
     
  8. Dec 4, 2015 #7
    okay and this?
    https://gyazo.com/39a967c89bee8109366e6f62991acf26 [Broken]
     
    Last edited by a moderator: May 7, 2017
  9. Dec 4, 2015 #8

    pwsnafu

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    Divide both sides by ##x^n## to get
    ## 1= \frac{x^{-n} - 1}{1-x}##
    Multiply both sides by 1-x
    ##1 - x = x^{-n} -1##
    Move everything to left
    ##0 = x^{-n} - 2 +x ##
     
  10. Dec 5, 2015 #9
    thank you very much
     
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