# Limit of arithmetic mean

## Homework Statement

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

## The Attempt at a Solution

i dont know abolutely. i tried definition

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

## The Attempt at a Solution

Related Calculus and Beyond Homework Help News on Phys.org
Hi Stanley!

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

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

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

$$|x_k-L|<\varepsilon$$

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

$$\frac{1}{n}\left|(x_1-L)+\cdots+(x_k-L)+\cdots+(x_n-L)\right|\le\frac{1}{n}\left|(x_1-L)+\cdots+(x_k-L)|+|x_{k+1}-L|+\cdots+(x_n-L)\right|$$

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

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 ??

What happens if n becomes bigger?

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

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