Proving Convergence of Averages of a Convergent Sequence

Thank you very much! I think I've got it now. In summary, to prove that the sequence (y_n) given by the averages of a convergent sequence (x_n) also converges to the same limit, we can use the inequalities |y_n - lim(x_n)| < e for an arbitrary e > 0 and the Archimedian property to bound the terms in the numerator of the first term of the sequence. We can also use the inequalities |x_n - lim(x_n)| < e for n > n_0 to bound the terms in the second term of the sequence.
  • #1
R.P.F.
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0

Homework Statement



Show that if (x_n) is a convergent sequence, then the sequence given by the averages

y_n = (x_1+ x_2 +...+ x_n)/n

also converges to the same limit.

Homework Equations





The Attempt at a Solution



I think I need to show that for n >= N, |y_n - lim(x_n)| < e for an arbitrary e > 0. I don't know where t start because y_n involves terms before N.
Any help is appreciated! :)
 
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  • #2
I'm not sure that this really works, but here's something you might try...

If you call x the limit of xn, you can estimate
[tex]|y_n - x| = \left| \frac{x_1}{n} + \frac{x_2}{n} + \cdots + \frac{x_n}{n} - x \right| \le \frac{1}{n} \left( |x_1 - x| + |x_2 - x| + \cdots |x_n - x| \right) [/tex]

At least then you'll have written everything in terms of |xi - x| which occurs in the limit definition of xn.

Another thought that occurred to me while writing this message is that you could take the maximum of the (finitely many) xi, and write
[tex] (x_1 + x_2 + \cdots + x_n) / n \le n x_i / n = x_i [/tex]
for some i = 1, 2, ..., n.
 
  • #3
A traditional (brute force in a sense) approach: Since [tex]x_n \to L[/tex], for every [tex]\epsilon>0[/tex] there exists [tex]n_0\in \mathbb{N}[/tex] such that for all [tex]n>n_0[/tex] we have [tex]L-\epsilon < x_n < L + \epsilon[/tex]

Then for [tex]n>n_0[/tex] we write:

[tex]\frac{ x_1 + x_2 + ... + x_n}{n} = \frac{ x_1 + x_2 + ... +x_{n_0}}{n} + \frac{x_{n_0 +1} + x_{n_0 +2} + ... + x_n}{n}[/tex]

For the first term, the numerator adds to some finite number, so we can use the Archimedian property to bound that, and for the second term we can use the inequalities above to bound that.
 
  • #4
Thank you so much, guys!
May I ask how to use LaTeX on PF? Thanks!
 
  • #5
If you click on the LaTex you can see the code that we typed to display what you see. Also there's a reference to a brief guide underneath the code. You'll pick it up by clicking on other peoples images and learning the basic structure of the code.
 
  • #6
Thank you, Gib. I know how to write LaTeX code. I'm just wondering how to post it onto PF. Do I just add [tex] and [\tex]?
 
  • #7
Yup just those before and after the code. Or for inline, [ itex] and [ /itex]
 

FAQ: Proving Convergence of Averages of a Convergent Sequence

How do you determine if a sequence is convergent or not?

To determine if a sequence is convergent, you need to look at the limit of the sequence. If the limit exists and is finite, then the sequence is convergent. If the limit does not exist or is infinite, then the sequence is divergent.

What is the definition of convergence of a sequence?

The definition of convergence of a sequence is that for any positive real number, there exists a positive integer N such that all terms in the sequence with an index greater than or equal to N are within that positive real number's distance from the limit of the sequence.

How do you prove the convergence of averages of a convergent sequence?

To prove the convergence of averages of a convergent sequence, you need to use the definition of convergence and show that for any positive real number, there exists a positive integer N such that all averages of the sequence with an index greater than or equal to N are within that positive real number's distance from the limit of the sequence.

Can a sequence be convergent but not have a convergent average?

Yes, a sequence can be convergent but not have a convergent average. This can happen if the sequence is oscillating around the limit, meaning that the average of the sequence does not approach a single value.

Why is proving convergence of averages of a convergent sequence important?

Proving convergence of averages of a convergent sequence is important because it allows us to use the limit of the sequence as an approximation for the average of the sequence. This can be useful in various mathematical and scientific applications where we need to estimate a value using a sequence of measurements or data points.

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