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Proving a Sequence is Convergent

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Let {[tex]a_{n}[/tex]}[tex]^{\infty}_{n=1}[/tex] be a sequence of real numbers that satisfies


    |[tex]a_{n+1}[/tex] - [tex]a_{n}[/tex]| [tex]\leq[/tex] [tex]\frac{1}{2}[/tex]|[tex]a_{n}[/tex] - [tex]a_{n-1}[/tex]|

    for all n[tex]\geq[/tex]2


    2. Relevant equations



    3. The attempt at a solution

    So, I know that it suffices to show that the sequence is Cauchy to prove this. And I can show that

    |[tex]a_{m}[/tex] - [tex]a_{n}[/tex]| [tex]\leq[/tex] [tex]\sum[/tex][tex]^{m-1}_{n}[/tex][tex]\frac{1}{2}[/tex][tex]^{k-1}[/tex] * |[tex]a_{2}[/tex] - [tex]a_{1}[/tex]|

    And that is about where I get lost. I'm very unclear on how to define the N in relation to the [tex]\epsilon[/tex].
     
  2. jcsd
  3. Oct 2, 2008 #2

    statdad

    User Avatar
    Homework Helper

    I will take your comment that you've shown

    [tex]
    |a_m - a_n | \le |a_2 - a_1| \sum_{k=n}^m \left(\frac 1 2 \right)^k
    [/tex]

    as correct - make sure it's correct.

    Suppose [tex] \varepsilon > 0 [/tex] is given. Try taking [tex] N [/tex] so large that [tex] n, m [/tex] larger than [tex] N [/tex] give

    [tex]
    \sum_{k=n}^m \left(\frac 1 2 \right)^k < \frac{\varepsilon}{2|x_2 - x_1|}
    [/tex]

    This should work (why the 2 in my denom - to make sure the sequence terms differ by less than [tex] \varepsilon [/tex] - it's a holdover from my student days, and I'm too old to change.)
     
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