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Math Help - Cauchy sequence

  1. #1
    Member SENTINEL4's Avatar
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    Cauchy sequence

    Hey guys i study for my exams at Wednesday and i would appreciate some help here.
    Based on "if for every ε>0, there is \nu_0\inN such that for all natural numbers m, n > \nu_o
    "

    1)Prove that the sequence a_\nu , \nu \in N , a_\nu=\frac{cos1}{1^2}+\frac{cos2}{2^2}+...+\frac{  cos\nu}{\nu} is a Cauchy sequence

    2) Prove that the sequence b_\nu , \nu \in N , b_\nu = 1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2\nu-1} is not a Cauchy sequence.

    Thanks in advance...
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  2. #2
    Senior Member BAdhi's Avatar
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    1)shouldn't it be \frac{cos1}{1^2}+\frac{cos2}{2^2}+...+\frac{cosv}{  v^2}

    cannot we use "if a sequence is covergent, it is a Cauchy sequence"
    Last edited by BAdhi; January 31st 2011 at 09:32 AM. Reason: changed the idea
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  3. #3
    Senior Member BAdhi's Avatar
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    sorry, posts keep repeating, something wrong with my connection
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  4. #4
    Member SENTINEL4's Avatar
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    Yes it is a_\nu=..........+\frac{cos\nu}{\nu^2}

    I am thinking something like :
    1) |a_{\nu+k}-a_\nu|=|(\frac{cos1}{1^2}+\frac{cos2}{2^2}+...+\fr  ac{cos(\nu+k)}{(\nu+k)^2})-(\frac{cos1}{1^2}+\frac{cos2}{2^2}+...+\frac{cos\n  u}{\nu^2})|=
    |\frac{cos(\nu+1)}{(\nu+1)^2}+\frac{cos(\nu+2)}{(\  nu+2)^2}+...+\frac{cos(\nu+k)}{(\nu+k)^2}|\leq|\fr  ac{cos(\nu+1)}{(\nu+1)^2}|+|\frac{cos(\nu+2)}{(\nu  +2)^2}|+...+|\frac{cos(\nu+k)}{(\nu+k)^2}|\leq
    \leq\frac{1}{(\nu+1)^2}+\frac{1}{(\nu+1)^2}+...+\f  rac{1}{(\nu+1)^2}\leq\frac{1}{\nu+1}+\frac{1}{\nu+  1}+...+\frac{1}{\nu+1}=\frac{k}{\nu+1}\rightarrow0

    2) |b_{2\nu}-b_\nu|=|(1+\frac{1}{3}+...+\frac{1}{2(2\nu)-1})-(1+\frac{1}{3}+...+\frac{1}{2\nu-1})|=|\frac{1}{2(\nu+1)-1}+\frac{1}{2(\nu+2)-1}+...+\frac{1}{2(2\nu)-1}|\geq
    \geq\frac{1}{4\nu-1}+\frac{1}{4\nu-1}+...+\frac{1}{4\nu-1}=\frac{\nu}{4\nu-1}=\frac{1}{4}
    so for \epsilon=\frac{1}{4}, b_\nu is not a Cauchy sequence.

    How is my work??
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