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Convergence of a sequence

  1. Sep 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that the sequence {xn}:
    xn := (21/1 - 1)2 + (21/2 - 1)2 + ... + (21/n - 1)2 is convergent.

    2. Relevant equations


    3. The attempt at a solution
    If n > m,
    |xn - xm| = (21/n - 1)2 + (21/(n-1) - 1)2 + ... + (21/(m+1) - 1)2
    < (21/n)2 + (21/(n-1))2 + ... + (21/(m+1))2
    < (21/(m+1))2 + (21/(m+2))2 + ...
    = 41/m
    Let ɛ > 0. We choose N such that 41/N < ɛ for all n > m > N.
    Then |xn - xm| < 41/N < ɛ for all n > m > N.
     
  2. jcsd
  3. Sep 20, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    Which N would you choose for ɛ=1?
     
  4. Sep 20, 2015 #3
    Just forgot it. I made a very stupid mistake.
    But I came up with the idea of letting bk = 21/k - 1.
    This means bk2 < 4[k(k-1)] for all k > 2.
    Therefore,
    xn = 1 + b22 + ... + bn2
    < 1 + 4/[1(2-1)] + ... 4/[n(n-1)]
    = 1 + 4(1 - 1/2) + ... + 4[1/(n-1) - 1/n]
    = 5 - 4/n
    < 5
    Since xn is monotone increasing and bounded, it is convergent.
    But is there any way to do it using the Cauchy criterion?
     
  5. Sep 20, 2015 #4

    mfb

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    2016 Award

    Staff: Mentor

    That step is certainly not trivial.

    I would use something like 21/k < 1 + c/k for some c.

    Cauchy criterion: Probably, but I don't see how it would help.
     
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