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: 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


    User Avatar
    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

    User Avatar
    Science Advisor
    Homework Helper

    Look at the new sequence ##r_n = a_n - L##.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted