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Subsequence convergence

  1. Oct 3, 2015 #1
    1. The problem statement, all variables and given/known data
    1. Consider the sequence $$\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5},\frac{1}{6}, \ldots$$ For which values ##z \in \mathbb{R}## is there a subsequence converging to ##z##?

    2. Prove that if ##\lim_{n\to \infty} x_n = z## then $$\lim_{n\to \infty} \frac{x_1+x_2+\ldots+x_n}{n} = z$$.

    2. Relevant equations


    3. The attempt at a solution
    1. If we take the subsequence ##\frac{1}{2}, \frac{2}{4}, \frac{3}{6}, \ldots## we can see that this is converging to ##\frac{1}{2} \in \mathbb{R}##. Am I on the right track or just not even close.

    2. No idea how to attack this one.

    Some help will be great thanks!!!!
     
  2. jcsd
  3. Oct 3, 2015 #2

    andrewkirk

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    For question 1:

    What is the range of values in the sequence?
    What does that tell you about the range of possible limits of subsequences?
    If any point in that second range were not a limit of a subsequence, what property would that point have to have, in relation to the sequence?
    Can any point in the range satisfy that property?
     
  4. Oct 3, 2015 #3

    Krylov

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    For question 2, you have to prove that
    $$
    C_n := \frac{1}{n}\sum_{k=1}^n{x_k} \to z \qquad \text{as } n \to \infty
    $$
    You could start by noting that this is equivalent to proving that
    $$
    \Bigl| C_n - z \Bigr| \to 0 \qquad \text{as } n \to \infty
    $$
    1. How can you arrange for ##z## to be included in the sum?
    2. Let ##\varepsilon > 0## be arbitrary. Why does there exist ##m \in \mathbb{N}## such that ##|x_k - z| \le \varepsilon## for all ##k \ge m##?
    3.You may now split the sum defining ##C_n## into two sums, assuming that ##n > m##. One of these sums vanishes in the limit ##n \to \infty## . (What property of convergent sequences do you use here?) In the same limit, the second sum is seen not to exceed ##\varepsilon##.
    4. Why is the argument completed?

    Incidentally, ##C_n## is often called the ##n##th partial Cesàro sum. What you will have shown is that ordinary convergence implies convergence of the partial Cesàro sums (also called "convergence in the mean") to the same limit. As an exercise, you could try to think about whether or not the converse is true as well, without looking it up of course.
     
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