1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook