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Cauchy Sequences mistake

  1. May 8, 2009 #1
    As far as I understand, a sequence converges if and only if it is Cauchy. So say for some sequence a_n and for all epsilon greater than zero we have [tex] |a_n - a_{n+1}| < \epsilon [/tex] for large enough n.

    We could then say a_n converges if and only if [tex] \lim_{n \rightarrow \infty} a_n - a_{n+1} = 0 [/tex].

    But what about if a_n = ln(n)?

    [tex] ln(n) - ln(n+1) = ln(n/(n+1)) [/tex] so for n tending to infinity ln(n) - ln(n+1) goes to 0. So I should be able to say that the sequence converges, but ln(n) obviously goes to infinity for increasing n.

    What's the mistake in my reasoning?
  2. jcsd
  3. May 8, 2009 #2
    The definition of Cauchy sequence does not just say [tex]
    |a_n - a_{n+1}| < \epsilon [/tex] . It says that for each [tex] \epsilon > 0 [/tex] there exists an integer N such that [tex] |a_n - a_m| < \epsilon [/tex] for all n,m > N. So, you can't just look at n and n+1. It's got to work for EVERY m and n greater than or equal to a certain integer N (which depends on epsilon). So, let's look at the case of ln(n). For each epsilon, is [tex] | ln (n) - ln (m) | = | ln(n/m| < \epsilon [/tex] for all m and n greater than a certain N?? No! This isn't true for every n and m. We can take n out far enough such that for any fixed epsilon, ln(n/m) is larger.
  4. May 9, 2009 #3
    Good explanation EbolaPox. JG89, one of the motivations for the definition of the Cauchy sequence is the idea that after a certain index N (n and m larger than N), any two terms of the sequence should be very close. This means that if we have a convergent sequence, we can do away with the limit, say L, and talk about the proximity of any two terms after a certain index.

    Having said that, you should prove the very first statement you made. You should be able to prove the direction convergent sequence implies Cauchy sequence without too much trouble as long as you keep in mind that there is nothing particularly special about the letter n that is usually used in forming the definition of convergence of a sequence {a_n}. The other direction involves a considerably larger amount of work, but the theorems typically used to prove Cauchy sequence implies convergent sequence: monotone convergence theorem and bolzano-weierstrass theorem have easy proofs if you can find them (the former is very intuitive, whereas the latter typically depends on a lemma in conjunction with the first theorem). On the other hand, if you have proved the nested intervals theorem and have possibly seen the proof of the IVT using nested intervals, there is a seemingly better motivated proof of the bolzano-weierstrass theorem which I could link you to if you were interested.
  5. May 9, 2009 #4
    Attention: This is completely wrong in full generality! By the very definition of a Banach space, this is only true in a Banach space.

    This is misleading. "The other direction" simply means you want to prove that the real numbers are a Banach space. Even if you forget about the construction of the reals, the theorems you suggest don't make sense in this context (Anyway, what do they have to do with the completeness of the real numbers? Monotone convergence?). There are nice ways of proving completeness of the reals.
  6. May 9, 2009 #5
    I judged what theorems are appropriate based on what I have seen from the OP's previous posts and experience. From my understanding, he is going through an introduction to analysis. While Cauchy sequences may be encountered frequently in classes that are more advanced than intro analysis, I am fairly confident I judged the context correctly with respect to the OP's understanding of the topics. However, I don't doubt that what you said was indeed simple, but as for misleading, perhaps you're right depending on the level of analysis the OP knows. Again, I think the OP is not concerned with demonstrating completeness of the reals, but rather working through sequences with epsilons and deltas, and there is no reason not to introduce Cauchy sequences in such a context; the proof that I outlined is much more appropriate if this is indeed the case.

    EDIT: Huh, well I tend to forget things at 4 AM. The outline for the other direction I had intended goes like this: show Cauchy sequences are bounded (typical take epsilon to be 1 or any positive number proof), use Bolzano-Weierstrass to get a convergent subsequence, and finally prove that if a subsequence of a Cauchy sequence converges, then the Cauchy sequence itself is convergent.

    Cliowa, I made the mistake of making it seem like the monotone convergence theorem is more important than it really is in this scenario. It is simply a theorem that can be used to prove the bolzano-weierstrass theorem for sequences.
    Last edited: May 10, 2009
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