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  1. #1
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    cauchy

    If 0<r<1 and for all n with natural numbers, show that is a Cauchy sequence
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by rmpatel5 View Post
    If 0<r<1 and for all n with natural numbers, show that is a Cauchy sequence
    A sequence (x_n) is a Cauchy sequence if :
    \forall ~ \varepsilon > 0, ~ \exists N \in \mathbb{N}, ~ \text{such that } \forall ~ p,q > N, ~ |x_p-x_q|< \varepsilon

    Note that since 0<r<1, \forall ~ \delta > 0, ~ \exists ~ N \in \mathbb{N}, ~ \text{such that } \forall ~ n > N, ~ r^n < \delta
    because the limit of r^n is 0.

    Now, write, assuming p>q, |x_p-x_q|=|(x_p-x_{p-1})+(x_{p-1}-\dots-x_{q+1})+(x_{q+1}-x_q)|.

    Use the triangle inequality : |x+y| \le |x|+|y|

    Try to rearrange all this information to answer the question
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    A sequence (x_n) is a Cauchy sequence if :
    \forall ~ \varepsilon > 0, ~ \exists N \in \mathbb{N}, ~ \text{such that } \forall ~ p,q > N, ~ |x_p-x_q|< \varepsilon

    Note that since 0<r<1, \forall ~ \delta > 0, ~ \exists ~ N \in \mathbb{N}, ~ \text{such that } \forall ~ n > N, ~ r^n < \delta
    because the limit of r^n is 0.

    Now, write, assuming p>q, |x_p-x_q|=|(x_p-x_{p-1})+(x_{p-1}-\dots-x_{q+1})+(x_{q+1}-x_q)|.

    Use the triangle inequality : |x+y| \le |x|+|y|

    Try to rearrange all this information to answer the question
    consider | x n+p - xn| = | xn+p - x n+p-1 + xn+p-1 -xn+p-2+xn+p-2-xn+p-3+xn+p-3- -------xn+1+xn+1-xn |
    = | xn+p - x n+p-1| + | xn+p-1 -xn+p-2|+ |xn+p-2-xn+p-3||+ | xn+p-3- ------+ |xn+2-xn+1| + | xn+1-xn |
    < r n+p-1 + r n+p-2 + rn+p-3 + --- + r n <
    we know that as n tends to ∞ , r n tends to 0.
    so, | x n+p - xn| < where = ε is as small as required depending on p.
    so, | x n+p - xn| < ε for all n,p > m where m depends on ε.
    so, { xn} is a cauchy sequence.

    right??
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