1. Not finding help here? Sign up for a free 30min 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!

Convergent sequence.

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that if [itex] x_n [/itex] is a convergent sequence, then the sequence given by that average values also converges to the same limit.
    [itex] y_n=\frac{x_1+x_2+x_3+.......x_n}{n} [/itex]
    3. The attempt at a solution
    Should I say that [itex] x_n [/itex] converges to some number P. so now I need to show that
    [itex] y_n [/itex] converges to P as well.
    Do I need to show that [itex] y_n-P< \epsilon [/itex]
     
  2. jcsd
  3. Feb 13, 2012 #2
    That's exactly what you need to show. Got any ideas?
     
  4. Feb 13, 2012 #3
    so [itex] x_n [/itex] is my nth term right, and also what the limit converges to.
    So [itex] x_n=P [/itex]
    So I should have[itex] (x_1+x_2+x_3+.......P)=Pn [/itex] maybe I can work on manipulating the sum and see if certain parts are less than other parts.
     
  5. Feb 13, 2012 #4
    Let me give you a hint:

    Define the limit of [itex]x_n[/itex] to be L then there is (for every epsilon>0) a N such that for [itex] n > N [/itex]

    [tex]|x_n-L|< \epsilon [/tex]

    Now we get:

    [tex] \left| \frac{x_1+x_2+x_3+...+x_n}{n} -\frac{nL}{n} \right| = \frac{|x_1-L| +|x_2-L| +...+|x_n-L|}{n} =\frac{|x_1-L| +|x_2-L| +...+|x_N-L| + |x_{N+1}-L|+...+|x_n-L|} {n} < \frac{|x_1-L| +|x_2-L| +...+|x_N-L|}{n} + \frac{(n-N)}{n} \epsilon < ...[/tex]

    Can you fill in the dots at the end?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Convergent sequence.
  1. Sequence Convergence (Replies: 6)

  2. Sequence Convergence (Replies: 2)

Loading...