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Question in my test today

  1. Dec 9, 2004 #1


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    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


    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


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    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


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    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


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    It's good enough. And if I could understand it, the others must have too.
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