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Convergence of a sequence

  1. Apr 21, 2008 #1
    This is related to my senior design project.

    I was wondering what would be a good way to check if a sequence is converging. As an example, imagine I have the following:

    for n = 1:10,000
    x(n) = f(n,n-1)

    What I'm trying to say is that the sequence is updated in each iteration, and that the value of each sequence element is a function of present and past inputs. I know that the sequence is converging. Lets say that it is converging to zero. However, the sequence may oscillate like crazy, and cross zero many times, but eventually it will settle.

    My question:
    How can I check if a sequence is converging to zero? What are some standard tests that I can look into?

    As of now, I have some hack code that basically computes the moving average. If the moving average gets to some threshold (i.e. not changing much), then I keep checking to see if it drops below some other threshold.
  2. jcsd
  3. Apr 22, 2008 #2


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  4. Apr 22, 2008 #3
    It would be worthwhile to pick up a book on analysis and read the chapter on sequences. You would be interested in the formal definition of convergence--

    [tex](a_n) \rightarrow A[/tex] if for any [tex]\epsilon > 0[/tex] there exists a natural number [tex]N[/tex] such that [tex]n\geq N \Rightarrow |a_n - A| < \epsilon[/tex].

    Another important thing you should know-- bounded, monotone sequences converge.

    Many times you can just straight up use the definition to confirm that a sequence converges. The trick is that many times you will set up a triangle inequality to get the job done. Which I even think is how Cauchy sequences (mentioned in the previous post) can be shown to converge.

    Let me rephrase that definition-- a sequence is convergent if for any interval around the proposed limit the sequence eventually enters the interval and never leaves. That means for any interval surrounding the proposed limit, there should be an infinite number of terms in the sequence contained within that interval.
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