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Math Help - showing its a cauchy sequence

  1. #1
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    showing its a cauchy sequence

    Show  i^n/n is a cauchy sequence. need  |(mi^n-ni^m)/nm| < e whenever n,m>_N .
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    Re: showing its a cauchy sequence

    Silly question, I suppose, but does i=\sqrt{-1} in this problem? If so, why not just use the triangle inequality?
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    MHF Contributor Drexel28's Avatar
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    Re: showing its a cauchy sequence

    The sequence converges, and so it's trivially Cauchy. It converges to zero because each term has modulus \frac{1}{n}.
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    Re: showing its a cauchy sequence

    true, but the nice thing about cauchy sequences is that there's no limits involved. note that:

    \left|\frac{i(m- n)}{mn}\right| = |i|\left|\frac{m-n}{mn}\right| = \left|\frac{m-n}{mn}\right|

     \leq \left|\frac{1}{n}\right| + \left|\frac{-1}{m}\right| = \left|\frac{1}{n}\right| + \left|\frac{1}{m}\right|

    if we choose N > 2/\epsilon, then for m,n > N

    \left|\frac{1}{n}\right| + \left|\frac{1}{m}\right| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon

    NOTE: if i represents a constant, rather than the square root of -1, then just choose:

    N > \frac{2|i|}{\epsilon}
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  5. #5
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    Re: showing its a cauchy sequence

    Deveno you misread slightly (it was i^m not mi) but the proof amounted to the same thing. Let e>0 be given.  |\frac{i^n}{n}-\frac{i^m}{m}|<_ |\frac{i^n}{n}|+|\frac{i^m}{m}|=\frac{1}{n}+\frac{  1}{m} Then choosing N >2/e completes the proof.
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  6. #6
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    Re: showing its a cauchy sequence

    indeed i did! good catch.
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