# Convergens of Cesáros mean

1. Nov 27, 2007

### Beowulf2007

Convergens of Cesáros mean(Urgent).

1. The problem statement, all variables and given/known data

I need to show the following:

(1) Let number sequence be given called $$\{\alpha_{n}\}_{n=1}^{\infty}$$ for which $$\alpha_{n} \rightarrow 0$$ where $$n \rightarrow \infty$$.

(2) Given a sequence $$\{\beta_{n}}\}_{n=1}^{\infty}$$ which is defined as $$\beta_{n}= \frac{\alpha_{1} + \alpha_{2} + \cdots + \alpha_{n}}{n}$$

If (1) is true then $$\beta_{n} \rightarrow 0$$ for $$n \rightarrow 0$$ is likewise true.

3. The attempt at a solution

The definition of convergens says (according to my textbook)

$$\forall \epsilon > 0 \exists N \in \mathbb{N} \forall n \in \mathbb{N}: n \geq N \Rightarrow |a_{n} - a| < \epsilon$$

If I use that on part 1 of (1) then

$$|\alpha_{n} - 0| < \epsilon$$ $$\forall n \in \mathbb{N}$$ thus $$\alpha_{n}$$ converges.

Regarding part (2).

I get the inequality $$|\beta_{n}| < \epsilon$$ from the definition.

Am I missing something here?

Best Regards
Beowulf

Last edited: Nov 27, 2007
2. Nov 27, 2007

### EnumaElish

The tricky part is, $\alpha_k$ can be > $\epsilon$ for k < N. (For example, $\alpha_1$ can be = 1,000,000$\epsilon$.) You need to argue "I can make n arbitrarily large, thus..."

3. Nov 27, 2007

### Beowulf2007

I can see if I make n large then epsilon will have to be very small. Is the point that epsilon can only be larger than $\alpha_n$ if n becomes extremly small?

Sincerely Yours
BeoWulf

4. Nov 27, 2007

### EnumaElish

You are starting with a given epsilon. The first k alphas may be larger than epsilon, so they'll make a large beta (if n were a constant). You have to see that, then argue: "this is not a problem, because I can make n arbitrarily large, which will make beta_n arbitrarily small."

$\epsilon > \alpha_n$ for n arbitrarily LARGE.

5. Nov 27, 2007

### Beowulf2007

So the that for any epsilon, if n is arbitrary larger then Beta_n will stay small and will therefore tend to zero? And that is simply the proof?

BR

Beowulf

6. Nov 27, 2007

### ZioX

Intuition: We are really taking finite means and then letting them go off to infinity. Now, eventually a_n becomes very close to zero and will always stay that close for any n afterwards (this is just convergence). But as we go off into infinity these a_ns get so close to zero, and there are so many of them, that they 'pull most of the mean' towards them.

Anyways, this is a tricky problem. But I'll start you off. Let epsilon>0. Since a_n->0 find N1 that works with epsilon/2. Then for all n>N1

b_n-0=a_1/n+a_2/n+...+a_N1/n + a_(N1+1)/n +...+a_n/n. Now the first N1 sums of b_n go to zero as n->infinity. Hence there exists an N2 such that whenever n>N2 |a_1/n+a_2/n+...+a_N1/n-0|<epsilon/2.

You should be able to finish this off. Man, I should have latexed this.