# Convergent sequence.

1. Feb 13, 2012

### cragar

1. The problem statement, all variables and given/known data
Show that if $x_n$ is a convergent sequence, then the sequence given by that average values also converges to the same limit.
$y_n=\frac{x_1+x_2+x_3+.......x_n}{n}$
3. The attempt at a solution
Should I say that $x_n$ converges to some number P. so now I need to show that
$y_n$ converges to P as well.
Do I need to show that $y_n-P< \epsilon$

2. Feb 13, 2012

### susskind_leon

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

3. Feb 13, 2012

### cragar

so $x_n$ is my nth term right, and also what the limit converges to.
So $x_n=P$
So I should have$(x_1+x_2+x_3+.......P)=Pn$ maybe I can work on manipulating the sum and see if certain parts are less than other parts.

4. Feb 13, 2012

### dirk_mec1

Let me give you a hint:

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

$$|x_n-L|< \epsilon$$

Now we get:

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

Can you fill in the dots at the end?