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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?
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