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Convergens of Cesáros mean

  1. Nov 27, 2007 #1
    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 [tex]\{\alpha_{n}\}_{n=1}^{\infty}[/tex] for which [tex]\alpha_{n} \rightarrow 0[/tex] where [tex]n \rightarrow \infty[/tex].

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

    If (1) is true then [tex] \beta_{n} \rightarrow 0[/tex] for [tex]n \rightarrow 0[/tex] is likewise true.

    3. The attempt at a solution

    The definition of convergens says (according to my textbook)

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

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

    [tex]|\alpha_{n} - 0| < \epsilon[/tex] [tex]\forall n \in \mathbb{N}[/tex] thus [tex]\alpha_{n}[/tex] converges.

    Regarding part (2).

    I get the inequality [tex]|\beta_{n}| < \epsilon[/tex] from the definition.

    Am I missing something here?

    Best Regards
    Beowulf
     
    Last edited: Nov 27, 2007
  2. jcsd
  3. Nov 27, 2007 #2

    EnumaElish

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    The tricky part is, [itex]\alpha_k[/itex] can be > [itex]\epsilon[/itex] for k < N. (For example, [itex]\alpha_1[/itex] can be = 1,000,000[itex]\epsilon[/itex].) You need to argue "I can make n arbitrarily large, thus..."
     
  4. Nov 27, 2007 #3
    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 [itex]\alpha_n[/itex] if n becomes extremly small?

    Sincerely Yours
    BeoWulf
     
  5. Nov 27, 2007 #4

    EnumaElish

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    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."

    [itex]\epsilon > \alpha_n[/itex] for n arbitrarily LARGE.
     
  6. Nov 27, 2007 #5
    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
     
  7. Nov 27, 2007 #6
    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.
     
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