Convergence of Cesáros Mean: Proving the Limit of a Sequence

[Beowulf2007]

Convergens of Cesáros mean(Urgent).

Homework Statement

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.

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

 

[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..."

 

[Beowulf2007]

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..."

Hi an Thank You for Your reply.

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 extremely small?

Sincerely Yours
BeoWulf

 

[EnumaElish]

I can see if I make n large then epsilon will have to be very small.

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."

Is the point that epsilon can only be larger than \alpha_n if n becomes extremely small?

\epsilon &gt; \alpha_n for n arbitrarily LARGE.

 

[Beowulf2007]

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 &gt; \alpha_n for n arbitrarily LARGE.

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

 

[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.