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Homework Help: Real Analysis: Stolz–Cesàro Proof

  1. Apr 28, 2010 #1

    Gib Z

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    1. The problem statement, all variables and given/known data
    1. Let xn and yn be sequences in R with yn+1 > yn > 0 for all natural numbers n and that yn→∞.
    (a) Let m be a natural number. Show that for n > m
    [tex]\frac{x_n}{y_n} = \frac{x_m}{y_n} + \frac{1}{y_n} \sum_{k=m+1}^{n} (x_k - x_{k-1})[/tex]

    (b) Deduce from (a) or otherwise that

    [tex]|\frac{x_n}{y_n}| \leq |\frac{x_m}{y_n}| + \sup_{k>m} | \frac{ x_k - x_{k-1} }{ y_k - y_{k-1} } |[/tex]

    (c) Assuming [tex]\frac{x_n-x_{n-1}}{y_n-y_{n-1}} \to 0[/tex], show [itex]x_n/y_n \to 0[/itex].

    (d) Assuming [tex]\frac{x_n-x_{n-1}}{y_n-y_{n-1}} \to L[/tex], show [itex]x_n/y_n \to L[/itex].

    2. Relevant equations
    N/A


    3. The attempt at a solution
    (a) was fine.

    (b) Would the question be more correct to use sup k>m+1 instead, since a k-1 index is in the inside expression?

    I'm not sure if this was the right thing to do as the question probably intended the sum to remain unsimplified, but I replaced it with xn-xm.

    Using the triangle inequality and that y is strictly increasing so that yn-ym < yn, we get

    [tex] | \frac{x_n}{y_n} | \leq | \frac{x_m}{y_n} | + | \frac{x_n-x_m}{y_n-y_m} | [/tex]

    Then I'm not really sure what I can validly do after that.

    (c) If (b) is assumed, then if we take the limit of both sides, it reduces to the statement that [tex] | \frac{x_n}{y_n} | - | \frac{x_m}{y_n} | \leq 0 [/tex] holds true for large n.

    I don't think that's the right direction to go, certainly since it seems to imply x is decreasing when that was never given. Don't know what to do.

    (d) No idea, but if I had (c) this one would probably be similar.
     
  2. jcsd
  3. Apr 29, 2010 #2

    Gib Z

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    Has anyone got any hints? I'm still stuck.
     
  4. Apr 29, 2010 #3


    you probably already know this, but the 2nd equation doesn't imply the first because sup(all that stuff) could be smaller than [tex] \left| \frac{x_n-x_m}{y_n-y_m} \right|[/tex]

    The most obvious path is to prove that
    [tex]\sup_{k>m}\left|y_n\frac{\Delta x_k}{\Delta y_k} \right| \geq \left| \sum_{k=m+1}^{n} \Delta x_k\right|[/tex]

    and then apply triangle inequality.


    First, restate it as an existence question, remove the sup and ask: does there exist a [tex]k[/tex] that makes the inequality true?

    [STRIKE]An idea:

    It reduces to finding [tex]m+1\leq k \leq n[/tex] such that

    [tex]\frac{y_n}{\Delta y_k}\geq n-m[/tex] (1)

    and

    [tex] \Delta x_k \geq \frac{|x_n-x_m|}{n-m} [/tex] (2)


    It is easy to to find a [tex]k[/tex] that satisfies each condition individually, but I can't see how to find one that satisfies both simultaneously =(
    [/STRIKE]
    edit: the above approach won't work, I found a counterexample to it.
    For c),


    to find N such that [tex]\frac{x_n}{y_n}<\varepsilon[/tex] for n>N,

    first find an m such that [tex]\frac{ \Delta x_k }{\Delta y_k } < \frac \varepsilon 2 [/tex] for k>m, then use part b) and finish it off.
     
    Last edited: Apr 29, 2010
  5. Apr 30, 2010 #4

    Gib Z

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    My mind started thinking clearly and I figured these out finally. If anyone's interested in the solutions send me a PM and I'll post them here.

    boboYO, I didn't see the edit until just now, my final solution for (c) didn't make any use of (b) and was quite long =[ Could you tell me how we could have used (b) to finish it off. When I see it I bet I'll kick myself.
     
    Last edited: Apr 30, 2010
  6. May 3, 2010 #5

    lanedance

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    see what you think of this
    let [itex] \Delta x_j = x_j-x_{j-1} [/itex]

    then, let k>m be the the supremum index, then
    [tex]| \frac{\Delta x_k }{ \Delta y_k }| \geq | \frac{\Delta x_j }{ \Delta y_j }|, \forall j>m [/tex]

    rearranging, knowing that [itex] \Delta y_j > 0 [/itex], then summing from m+1 upto n:
    [tex]| \frac{\Delta x_k }{ \Delta y_k }| \sum \Delta y_j \geq \sum |\Delta x_j |\geq |\sum \Delta x_j| [/tex]
    so
    [tex]\sup_{k>m}| \frac{\Delta x_k }{ \Delta y_k }| \geq |\frac{x_n-x_m}{y_n-y_m}|\geq |\frac{x_n-x_m}{y_n}| \geq |\frac{x_n}{y_n}| - |\frac{x_m}{y_n}| [/tex]
     
  7. May 3, 2010 #6

    lanedance

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    then for c) choose M such that
    [tex]\frac{x_m-x_{m-1}}{y_m-y_{m-1}} < \epsilon, \forall m>M [/tex]

    then for n,m>M
    [tex]|\frac{x_n}{y_n}| \leq |\frac{x_m}{y_n}| + \epsilon[/tex]

    [itex] x_m [/itex] is just a number, and as [itex] y_n [/itex] is unbounded as n increases, we should be able to choose N>M so that for all n>N
    [tex] |\frac{x_m}{y_n}|\leq \epsilon[/tex]

    then for n>N>M
    [tex]|\frac{x_n}{y_n}| \leq 2\epsilon[/tex]

    which should be pretty close....


    this part threw me for a bit above
    [tex]\sup_{k>m}| \frac{\Delta x_k }{ \Delta y_k }| \geq |\frac{x_n-x_m}{y_n-y_m}| [/tex]
    but i'm thinking maybe the continuous analogue is something like the mean value theorem, except that the maximum gradient must be >= the average

    not 100% there is no holes...
     
    Last edited: May 3, 2010
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