What is the Limit of the Arithmetic Mean as n Approaches Infinity?

[stanley.st]

Homework Statement

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

Homework Equations

The Attempt at a Solution

i don't know abolutely. i tried definition

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

 

[micromass]

Hi Stanley!

First, what does it mean that \lim_{n\rightarrow +\infty}{x_n}=L?? I simply want the definition...

 

[stanley.st]

Thanks for reply.

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

I'm not sure about right steps. I can't simply write

|x_k-L|<\varepsilon

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

\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|

 

[micromass]

Yes, so take that N such that |x_n-L|<\epsilon.

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

Now, let me do the first few steps:

\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|}

Now try to go on

 

[stanley.st]

Then I should write

<\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}

Terms in first sum I can bound by maximum

<\frac{N}{n}max+\frac{n-N}{n}\epsilon

And then ??

 

[micromass]

What happens if n becomes bigger?

Can you find an n such that the entire sum becomes smaller then \epsilon? (or rather 2\epsilon?)

 

[stanley.st]

OMG, I'm so stupid. Thank you so much man.