Results 1 to 2 of 2

Thread: sequence, convergence

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    16

    sequence, convergence

    Suppose that $\displaystyle (x_n)$ is a sequence in $\displaystyle \mathbb{R}$. Define a sequence $\displaystyle (y_n)$ by $\displaystyle y_n=\frac{x_1+x_2+\cdots+x_n}{n}$, $\displaystyle \forall n \in \mathbb{N}$. Prove that if $\displaystyle (x_n)$ converges to $\displaystyle x \in \mathbb{R}$ then $\displaystyle (y_n)$ converges to $\displaystyle x$.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by xboxlive89128 View Post
    Suppose that $\displaystyle (x_n)$ is a sequence in $\displaystyle \mathbb{R}$. Define a sequence $\displaystyle (y_n)$ by $\displaystyle y_n=\frac{x_1+x_2+\cdots+x_n}{n}$, $\displaystyle \forall n \in \mathbb{N}$. Prove that if $\displaystyle (x_n)$ converges to $\displaystyle x \in \mathbb{R}$ then $\displaystyle (y_n)$ converges to $\displaystyle x$.
    This is Cesaro's mean. And you want to prove Cesaro's lemma.
    There is a proof in the French Wikipedia (Lemme de CesÓro - WikipÚdia), but not in the English one.
    I'll translate and adapt it here...


    $\displaystyle x_n \to x$ can be translated this way :
    $\displaystyle \forall \varepsilon >0,~ \exists N \in \mathbb{N},~ \forall n>N,~ |x_n-x|< \varepsilon/2$

    Let $\displaystyle S_N=\sum_{k=1}^N |x_k-x|$

    Let's assume n>N :
    $\displaystyle |y_n-x|=\left|\frac 1n \sum_{k=1}^n (x_k-x)\right|$
    By the triangle inequality, we have :
    $\displaystyle |y_n-x| \leqslant \frac 1n \sum_{k=1}^N |x_k-x|+\frac 1n \sum_{k=N+1}^n |x_k-x|$
    $\displaystyle |y_n-x| \leqslant \frac{S_N}{n}+\frac{n-N}{n} (\varepsilon/2)=\frac{S_N}{n}+\varepsilon/2 -\frac{N}{n} (\varepsilon/2) \leqslant \frac{S_N}{n}+\varepsilon/2$
    (because N/n (epsilon/2) >0)

    But $\displaystyle S_N$ doesn't depend on n. So $\displaystyle \lim_{n \to \infty} \frac{S_N}{n}=0$
    This means that for any $\displaystyle \varepsilon >0$, there exists an integer N' such that for all n>N', we have :
    $\displaystyle \left|\frac{S_N}{n}\right|\leqslant \varepsilon/2$


    By combining this latter inequality to the previous one, we have, for $\displaystyle n>\max\{N,N'\}$ :
    $\displaystyle |y_n-x| \leqslant \varepsilon/2+\varepsilon/2=\varepsilon \qquad \square$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. convergence of a sequence
    Posted in the Differential Geometry Forum
    Replies: 9
    Last Post: Mar 24th 2011, 03:11 PM
  2. Replies: 7
    Last Post: Oct 12th 2009, 10:10 AM
  3. Replies: 6
    Last Post: Oct 1st 2009, 09:10 AM
  4. Convergence of sequence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Sep 23rd 2009, 05:33 AM
  5. Replies: 6
    Last Post: Oct 24th 2008, 01:45 PM

Search Tags


/mathhelpforum @mathhelpforum