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Cauchy Sequence

  1. Apr 5, 2013 #1
    I am looking for a different proof that ##(S_n) = \frac{1}{n}## is cauchy.

    The regular proof goes like this (concisely):

    ##\left|\frac{1}{n} - \frac{1}{m} \right| \leqslant \left|\frac{m}{nm}\right| \ (etc...) \ <\epsilon ##

    but I was thinking about an alternative proof. Is my proof correct:

    let ##\epsilon > 0## by Archimedian property ##\exists N \ s.t. \frac{1}{N}<\epsilon##

    This is equivalent to ##\frac{1}{N}<\frac{\epsilon}{2}## "may be ommited"

    Now ##\forall n, m \geqslant N## we have by ##\Delta## ineq.

    ##\left|\frac{1}{n} - \frac{1}{m} \right| \leqslant \left|\frac{1}{n} \right| + \left|\frac{1}{m} \right|\leqslant \frac{1}{N}+\frac{1}{N} < ε##

    What do you guys think? Thanks...
     
  2. jcsd
  3. Apr 5, 2013 #2
    I think it's perfectly ok.
     
  4. Apr 6, 2013 #3

    jbunniii

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    I'm not sure what this means, but the rest is fine.

    Another way to prove it is to note that ##(s_n)## converges to ##0## (easy proof), and use the fact that any convergent sequence is Cauchy.

    Proof: if ##(x_n)## is a sequence which converges to some number ##L##, then given ##\epsilon > 0##, there is some ##N## for which ##|x_n - L| < \epsilon / 2## for all ##n \geq N##. Therefore, if ##m \geq N## and ##n \geq N##, then ##|x_n - x_m| = |x_n - L + L - x_m| \leq |x_n - L| + |x_m - L| \leq \epsilon##.
     
  5. Apr 6, 2013 #4
    Well since ε in this case can be so small, then we can use a larger N > 1/ε (actually twice larger) to end up with ε/2+ε/2 in the end instead of 2ε.
     
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