Limit of arithmetic mean

  • Thread starter stanley.st
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  • #1
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Homework Statement


prove: lim x_n = L. Then
[tex]\lim_{n\to\infty}\frac{x_1+\cdots+x_n}{n}=L[/tex]


Homework Equations





The Attempt at a Solution



i dont know abolutely. i tried definition

[tex]\left|\frac{x_1+\cdots+x_n}{n}-L\right|=\frac{1}{n}\left|(x_1-L)+\cdots+(x_n-L)\right|[/tex]

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
22,089
3,286
Hi Stanley!

First, what does it mean that [tex]\lim_{n\rightarrow +\infty}{x_n}=L[/tex]?? I simply want the definition...
 
  • #3
31
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Thanks for reply.

[tex]\forall\varepsilon>0\exists N\in\mathbb{N}\forall n>N:|x_n-L|<\varepsilon[/tex]

I'm not sure about right steps. I cant simply write

[tex]|x_k-L|<\varepsilon[/tex]

for some k in that sum. I should divide this sum into two parts

[tex]\frac{1}{n}\left|(x_1-L)+\cdots+(x_k-L)+\cdots+(x_n-L)\right|\le\frac{1}{n}\left|(x_1-L)+\cdots+(x_k-L)|+|x_{k+1}-L|+\cdots+(x_n-L)\right|[/tex]
 
  • #4
22,089
3,286
Yes, so take that N such that [tex]|x_n-L|<\epsilon[/tex].

Now, our goal is to make [tex]|(x_1+...+x_n)/n-L|[/tex] smaller then epsilon.

Now, let me do the first few steps:

[tex]\left|\frac{1}{n}\sum_{k=1}^n{x_k}-L\right|\leq \sum_{k=1}^n{\frac{1}{n}|x_k-L|}=\sum_{k=1}^N{\frac{1}{n}|x_k-L|}+\sum_{k=N+1}^n{\frac{1}{n}|x_k-L|}[/tex]

Now try to go on
 
  • #5
31
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Then I should write

[tex]<\sum_{k=1}^N{\frac{1}{n}|x_k-L|}+\sum_{k=N+1}^n{\frac{1}{n}\epsilon}=\sum_{k=1}^N{\frac{1}{n}|x_k-L|}+{\frac{n-N}{n}\epsilon}[/tex]

Terms in first sum I can bound by maximum

[tex]<\frac{N}{n}max+\frac{n-N}{n}\epsilon[/tex]

And then ??
 
  • #6
22,089
3,286
What happens if n becomes bigger?

Can you find an n such that the entire sum becomes smaller then [tex]\epsilon[/tex]? (or rather [tex]2\epsilon[/tex]?)
 
  • #7
31
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OMG, I'm so stupid. Thank you so much man.
 

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