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Prove a Cauchy Sequence using Geometric Sums

  1. Apr 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Let {x_n} be a sequence. and let r be a real number 0<r<1. Suppose |x_(n+1) - x_n|<=r|x_n -x_(n-1)| for all n>1. Prove that {x_n} is Cauchy and hence convergent.

    2. Relevant equations

    if |r|<1 then the sequence [tex]\sum r^k[/tex] from k=0 to n converges to 1/(1-r)

    3. The attempt at a solution

    If I let {x_n}=r^2+r^3+...+r^n &
    {x_(n+1)}=r^3+r^4+...+r^(n+1) &
    I can then take |x_(n+1) - x_n|=|r^(n+1)-r^2|
    and |x_n -x_(n-1)|=|r^n-r|

    If I plug these into my given inequality I have |r^(n+1)-r^2|<=r||r^n-r|

    I can then say that since we know 0<r<1 and from our relevat equations we know [tex]\sumr^k[/tex] converges then we know {x_n} is Cauchy and hence converges since every convergent sequence of real numbers is Cauchy.

    I am not sure.
  2. jcsd
  3. Apr 17, 2009 #2
    Use this
    |x_(n+1) - x_n|<=r|x_n -x_(n-1)|
    to show that

    |x_(n+1) - x_n| <= r^(n-1)|x_2 - x_1|
    which can be done by induction

    By the way,
    You cannot say this:
    x_n=r^2+r^3+...+r^n &
    x_(n+1)=r^3+r^4+...+r^(n+1) &

    x_n is an ARBITRARY sequence with the property that |x_(n+1) - x_n|<=r|x_n -x_(n-1)|.
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