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Proving a sequence is Cauchy

  1. Jun 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Proposition19part1-1.png


    2. Relevant equations

    See picture in 1.

    3. The attempt at a solution

    See picture in 1.

    I think what's tripping me up is that I'm not sure how to go about picking my N. I want to show the sequence has a limit by proving that it is Cauchy.
     
    Last edited: Jun 7, 2009
  2. jcsd
  3. Jun 7, 2009 #2

    Dick

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    That's really not at all convincing. You know the limit is zero and the sequence is decreasing, right? Why can't you look at the limit of log(p^n)=n*log(p)?
     
  4. Jun 7, 2009 #3
    I do not know the limit is zero. That is the next portion of the proof. I only wish to prove the limit exists by using the definition of a Cauchy sequence. Because we know that any sequence which is Cauchy is convergent.
     
  5. Jun 7, 2009 #4

    Dick

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    Ok, but I still don't see why you can't use p^n=exp(log(p)*n) to show it converges to 0 and that it is Cauchy at the same time. I.e. find an N such that p^N<epsilon and use that 0<p^n<p^N for n>N.
     
    Last edited: Jun 7, 2009
  6. Jun 7, 2009 #5
    Well generally to prove any sort of limit (of a sequence, of a function, etc.), you really should have an idea of what the limit actually is. As dick mentioned, the limit is clearly 0 since if 0 < p < 1, then as n gets very large, p^n tends to zero (and naturally, the N you choose in the definition of convergent or cauchy sequences will be bigger for p closer to 1, but < 1). Figuring these parts out helps tremendously in checking the eventual proof by the formal definition.

    Now in this case, it seems even easier to actually prove that the sequence converges to 0, but assuming you did not know this, the following should work.

    We need |p^m - p^n| < epsilon for m, n > N. Clearly, p^m > 0 (since p > 0). Moreover, note that if n > N, then p^n < p^N (Why?). You should be able to complete the proof now.
     
  7. Jun 7, 2009 #6
    "We need |p^m - p^n| < epsilon for m, n > N. Clearly, p^m > 0 (since p > 0). Moreover, note that if n > N, then p^n < p^N (Why?) Because 0 < p < 1, so raising it to a higher exponent makes it smaller. You should be able to complete the proof now."

    But hold on snipez90, what is the big "N" you mentioned here? Where did it come from? How did you pick it, because one of the things I wish to prove is that there is such an N.
     
  8. Jun 7, 2009 #7

    Dick

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    More abstractly, you have a sequence that's decreasing and bounded below. Look at the greatest lower bound. It's always Cauchy. Use the definition of 'greatest lower bound' and the monotonicity of the sequence. To show the greatest lower bound is really zero, use the log.
     
  9. Jun 8, 2009 #8
    Or you could use the inequality: [tex] (1 + h)^n \ge 1 + hn [/tex] where n is an integer and h is a positive number. Since 0 < p < 1, set p = 1/(1+h).

    Use the inequality to find an upper bound for the sequence that converges to 0.
     
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