Proving Convergence: Showing That x_n and y_n Have the Same Limit

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cragar
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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]
 
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That's exactly what you need to show. Got any ideas?
 
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.
 
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?