Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question in my test today

  1. Dec 9, 2004 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There was this question in my analysis exam today. I have a feeling it should be easy but no one I've asked knew how to do it.

    We associate to the sequence [itex]\{a_n\}[/itex] the sequence defined by

    [tex]b_n=\frac{a_1+a_2+...+a_n}{n}[/tex]

    Show that if [itex]\{a_n\}[/itex] converges towards a, then [itex]\{b_n\}[/itex] converges towards a.

    I realised that

    [tex]b_n=\frac{\sum_{n=1}^{\infty} a_n}{n}[/tex]

    or even

    [tex]b_n=\frac{\sum_{k=1}^{n} a_k}{\sum_{k=1}^{n} 1}[/tex]

    but all my attemps involving epsilon-delta, convergence tests, Cauchy convergence "caracterisation", etc. failed. Please tell me how to do this. Thanks a lot.
     
  2. jcsd
  3. Dec 9, 2004 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It often helps to split convergent sequences into two parts, an initial part, and the rest, because you can decide how much "the rest" resembles the limit.
     
  4. Dec 9, 2004 #3
    let [tex] \epsilon > 0 [/tex]. then since a_n converges there exists N>0 such that [tex] a - \epsilon < a_n < a + \epsilon [/tex].

    now [tex] b_n = \frac{a_1 + a_2 + ... + a_n}{n}[/tex] for n[tex]\geq[/tex]

    so [tex] b_n = \frac{a_1 + a_2 + ... + a_N}{n} + \frac{a_1 + a_2 + ... + a_n}{n}[/tex]

    & since [tex] \frac{(n-N)(a-\epsilon)}{n} < \frac{a_N+1 + ... + a_n}{n} < \frac{(n-N)(a+\epsilon)}{n}[/tex],

    [tex] \frac{a_1 + a_2 + ... + a_N}{n} + \frac{(n-N)(a-\epsilon)}{n} < b_n < \frac{a_1 + a_2 + ... + a_N}{n} + \frac{(n-N)(a+\epsilon)}{n} [/tex]

    now using that fact that [tex]\limsup a_n \leq \limsup b_n[/tex] if [tex]a_n \leq b_n[/tex] on the previous inequality we get

    [tex]\limsup ( \frac{a_1 + a_2 + ... + a_N}{n} + \frac{(n-N)(a-\epsilon)}{n} ) \leq \limsup b_n \leq \limsup \frac{a_1 + a_2 + ... + a_N}{n} + \frac{(n-N)(a+\epsilon)}{n}[/tex].

    since [tex]\lim ( \frac{a_1 + a_2 + ... + a_N}{n} + \frac{(n-N)(a-\epsilon)}{n} ) = 0 + (a - \epsilon[/tex] )
    and [tex]\lim ( \frac{a_1 + a_2 + ... + a_N}{n} + \frac{(n-N)(a+\epsilon)}{n} ) = 0 + (a + \epsilon[/tex] )

    we can conclude that
    a - [tex]\epsilon \leq \limsup b_n \leq a+\epsilon[/tex], for any [tex]\epsilon > 0[/tex]

    since epsilon is arbitrary, [tex] a \leq \limsup b_n \leq a [/tex], so [tex] a = \limsup b_n[/tex], similarily for liminf, and the result follows from the limsup & liminf being equal
     
    Last edited: Dec 9, 2004
  5. Dec 9, 2004 #4

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    holy cow!

    I now feel my feeling was injustified. It's a beautiful proof and very instructive; a gazilion thanx fourier!
     
  6. Dec 10, 2004 #5
    there are a couple lines that need fixing but i wonder if it would be worth the trouble or if people can still follow it. i need to get used to the tex-ing
     
  7. Dec 10, 2004 #6

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's good enough. And if I could understand it, the others must have too.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question in my test today
  1. Was my test wrong? (Replies: 4)

  2. Problem on test today (Replies: 2)

  3. Today test - one limit (Replies: 3)

Loading...