Convergent sequence.

  • Thread starter cragar
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  • #1
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Homework Statement


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]

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]
 

Answers and Replies

  • #2
That's exactly what you need to show. Got any ideas?
 
  • #3
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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.
 
  • #4
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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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