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Sequence Convergence

  1. Mar 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider the sequence {an}[itex]\subset[/itex]R which is recursively defined by an+1=f(an). Prove that if there is some L[itex]\in[/itex]R and a 0≤c<1 such that |[itex]\frac{a_{n+1}-L}{a_{n}-L}[/itex]|<c for all n[itex]\in[/itex]N then limn[itex]\rightarrow[/itex]∞an=L.


    2. Relevant equations
    Definition of convergence: Suppose (X,d) is a metric space and let {a_{n}}[itex]\subseteq[/itex]R denote a sequence in R. We will say {a[itex]_{n}[/itex]} converges in R if there is an L[itex]\in[/itex]R such that for every ε>0 there is an N'[itex]\in[/itex]N so that n'≥N' implies that d(a[itex]_{n}[/itex],L)<ε [itex]\Rightarrow[/itex] limn[itex]\rightarrow[/itex]∞an=L.


    3. The attempt at a solution
    Take ε>0. Since |an+1-L|< |an-L| [itex]\Rightarrow[/itex] |[itex]\frac{a_{n+1}-L}{a_{n}-L}[/itex]|<c. If n'=n+1 then N'=n. If 0<ε<|an+1-L| then d(an,L)=|an-L|<ε which by definition implies limn[itex]\rightarrow[/itex]∞an=L.

    I'm not really sure if this is right. If anyone could tell me if anything is wrong with it that would be great! Thanks.
     
  2. jcsd
  3. Mar 29, 2014 #2

    pasmith

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    Homework Helper

    You can give the definition of convergence of a sequence in a general metric space (X,d) and then immediately tell us that [itex]X = \mathbb{R}[/itex] and [itex]d: (x,y) \mapsto |x - y|[/itex], or you could just give the definition of convergence of a real sequence:

    A real sequence [itex](a_n)[/itex] converges to [itex]L \in \mathbb{R}[/itex] if and only if for all [itex]\epsilon > 0[/itex] there exists an [itex]N \in \mathbb{N}[/itex] such that for all [itex]n \in \mathbb{N}[/itex] if [itex]n \geq N[/itex] then [itex]|a_n -L| < \epsilon[/itex] (or, equivalently, [itex]L - \epsilon < a_n < L + \epsilon[/itex]).

    You are given [itex]\left|\dfrac{a_{n+1}-L}{a_{n}-L}\right| < c < 1[/itex] for all [itex]n \in \mathbb{N}[/itex]. You conclude that [itex]|a_{n+1} - L| < |a_n - L|[/itex] for all [itex]n \in \mathbb{N}[/itex].

    This tells you that [itex]|a_n - L|[/itex] is a strictly decreasing sequence bounded below by zero, and that it therefore converges to some [itex]S \geq 0[/itex].

    You now need to show that [itex]S > 0[/itex] is impossible.
     
  4. Mar 29, 2014 #3

    Ray Vickson

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    Homework Helper

    Look at the new sequence ##r_n = a_n - L##.
     
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