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!

Real analysis: Sequences question

  1. Apr 13, 2009 #1
    1. The problem statement, all variables and given/known data
    If Xn is bounded by 2, and [tex]|X_{n+2} - X_{n+1}| \leq \frac{|X^2_{n+1} - X^2_n|}{8} [/tex], prove that Xn is a convergent sequence.


    2. Relevant equations



    3. The attempt at a solution
    I believe the solution lies in proving Xn a Cauchy sequence, but I'm not sure how to work it out. I considered |Xn - Xm| adding and subtracting the terms between m and n but i got stuck.
    I also tried to check for telescoping, with no luck.
     
    Last edited: Apr 13, 2009
  2. jcsd
  3. Apr 13, 2009 #2
    Factor [tex]X^2_{n+1} - X^2_n[/tex].

    Then get an upper bound for [tex]\frac{|X^2_{n+1} - X^2_n|}{8}[/tex].
     
  4. Apr 13, 2009 #3
    [tex]
    \frac{|X^2_{n+1} - X^2_n|}{8} = \frac{|(X_{n+1} + X_n)(X_{n+1} - X_n)|}{8} \leq \frac{|X_{n+1} - X_n|}{2}
    [/tex]

    Iterating, I reached [tex] |X_{n+2} - X_ {n+1}| \leq \frac{|X_2 - X_1|}{2^n} [/tex]
    I'm not sure if I continued in the right track.. but I'm stuck.
    Thanks for your input.
     
  5. Apr 13, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Now go back to trying to show the sequence is Cauchy by adding and subtracting terms in |xn-xm|. Use the triangle inequality.
     
  6. Apr 13, 2009 #5
    Or could even look at the sequence of partial sums, getting it bounded above.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Real analysis: Sequences question
Loading...