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Recursive sequences convergence

  1. Mar 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Let the sequence {a_n} defined by:

    a_n+1 = a_n/[sqrt(0.5a_n + 1) + 1]

    Prove that {a_n} converges to 0

    2. Relevant equations

    3. The attempt at a solution

    I tried manipulating the equation but to no avail....
  2. jcsd
  3. Mar 1, 2008 #2
    If the sequence approaches 0, then lim(n->infinity) a_n should be 0 as should a_n+1 correct?

    If the series converges, then a_n and a_n+1 should both approach the limit L of the sequence as n becomes large.
    Last edited: Mar 1, 2008
  4. Mar 1, 2008 #3
    yes. but i dont see how that helps
  5. Mar 1, 2008 #4
    Ok let me try it a different way. Suppose instead of explicitly labeling the sequence as a sequence, I say that the sequence is actually a function f(x) which takes x_n and transforms it into x_n+1.

    For this sequence, [itex]f(x)=\frac{x}{\sqrt{x/2+1}+1}[/itex]. If the sequence converges to a limit L, then clearly f(L)=L. Does that make more sense?
    Last edited: Mar 1, 2008
  6. Mar 1, 2008 #5
    yes, i can see how that shows that the limit is 0, but i don't think its rigorous enough.
    is there some way to actually show that the limit is 0, perhaps using definitions of limits?
  7. Mar 1, 2008 #6
    I had to delete my previous post because I found an error in it, but I think the argument can be made by comparison. Using the definition of a limit, for every [tex]\epsilon > 0[/tex], you want to prove there exists an N such that for [tex]n > N[/tex],[tex]|x_n - L| < \epsilon[/tex].

    The question is vague enough and seems to ignore the lack of convergence if [tex]x_n = -2[/tex] so I'm going to restrict my argument to [tex]x_n > 0[/tex]. Clearly for arbitrary w and u > 0 such that w > u, then [tex]\frac{1}{\sqrt{0.5w+1}+1} < \frac{1}{\sqrt{0.5u+1}+1}[/tex]. Using this, you can place an upper bound on the ratio of subsequent terms of the sequence [tex]x_{n+1} = \frac{x_n}{\sqrt{0.5x_n+1}+1}[/tex] to say that [tex]x_{n+1} \leq \frac{1}{2} x_n[/tex], which is much easier to work with.

    I don't think attempts to directly apply the definition of a limit without using this kind of workaround would be met with much success.
    Last edited: Mar 1, 2008
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