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Cauchy sequence question

  1. Oct 16, 2008 #1
    Let a_1 and a_2 be arbitrary real number that are not equal. For [tex]n \geq 3[/tex], define a_n inductively by,

    [tex]a_n = \frac{1}{2} (a_{n-1}+a_{n+2} )[/tex]



    I cannot get the result that the book gets. I proceed,

    [tex]a_{n+1} - a_{n} = \frac{1}{2}(a_n + a_{n-1} ) - \frac{1}{2} (a_{n-1} + a_{n-2} ) = \frac{1}{2} ( a_n - a_{n-2}
    )= \frac{1}{2}(a_n + a_{n-1} ) [/tex]

    The book got the answer,

    [tex] a_{n+1} - a_n = \frac{-1}{2} (a_n - a_{n-1} ) [/tex]

    Any help for me?
     
  2. jcsd
  3. Oct 16, 2008 #2

    Dick

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    You've got some typos in there that are making this really confusing. The original recursion is a_n=(1/2)*(a_n-1+a_n-2), right? That says 2*a_n-a_n-1=a_n-2. I agree with the book.
     
  4. Oct 16, 2008 #3
    I can't see it. I've done some algebra and I still can't get anywhere from your last step.
     
  5. Oct 16, 2008 #4

    Dick

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    You've got a_n+1-a_n=(1/2)(a_n-a_n-2), right? Put the expression for a_n-2 into that. If you still aren't getting it show your work.
     
  6. Oct 16, 2008 #5
    I got it. Thanks Dick
     
  7. Oct 17, 2008 #6

    HallsofIvy

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    Here's a "cheat":
    Since [itex]\sum x^n/n![/itex] is the Taylor's series for ex, 1+ 1/2!+ 1/3!+ ... converges to e. Since it converges, it is a Cauchy sequence!

    Or you could just note that the numeric series converges by the ratio test. Again, since it converges, it is a Cauchy sequence.

    But I doubt that either of those answers is what was wanted!
     
  8. Oct 17, 2008 #7
    Those are nice things to put the problem into perspective though. Sometimes I (and hopefully others) get lost in the algebra of the problem.
     
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