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Math Help - limit question

  1. #1
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    limit question

    Here goes: I know that the limit as n goes to infinity of (
    an) = A. I need to find the limit of the following as n goes to infinity:
    (a1 + 2a2 + 3a3 + · · · + nan)
    ---------------------------------------
    [n(n+1)] / 2

    OK. I know the limit goes to A. And I know that (1 + 2 + 3 + ... + n) = [n(n+1)] / 2. So I want to say that this limit equals
    (1 + 2 + 3 + ... + n)*(an)
    ------------------------, which would be just (an) and therefore its limit is A
    [n(n+1)] / 2

    Although I'm not sure if this works, since I would be multiplying each coefficient by an and not by a1, a2, or whatever.

    Is there a different way to show this?

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  2. #2
    MHF Contributor red_dog's Avatar
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    Apply Stolz-Cesaro theorem: Stolz-CesÃ*ro theorem - Wikipedia, the free encyclopedia

    Let x_n=1\cdot a_1+2\cdot a_2+\ldots+na_n, \ y_n=1+2+\ldots+n.
    (y_n) is a sequence with positive terms, strictly ascending and unbounded.

    Then \displaystyle\lim_{n\to\infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n}=\lim_{n\to\infty}\frac{(n+1)a_{n+1}}{n+1}=\li  m_{n\to\infty}a_{n+1}=A.

    So, \displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}=A
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by BrainMan View Post
    Here goes: I know that the limit as n goes to infinity of (
    an) = A. I need to find the limit of the following as n goes to infinity:
    (a1 + 2a2 + 3a3 + · · · + nan)
    ---------------------------------------
    [n(n+1)] / 2

    OK. I know the limit goes to A. And I know that (1 + 2 + 3 + ... + n) = [n(n+1)] / 2. So I want to say that this limit equals
    (1 + 2 + 3 + ... + n)*(an)
    ------------------------, which would be just (an) and therefore its limit is A
    [n(n+1)] / 2

    Although I'm not sure if this works, since I would be multiplying each coefficient by an and not by a1, a2, or whatever.

    Is there a different way to show this?
    Well, a_1 + 2a_2 + ~ ... ~ + na_n \neq (1 + 2 + ~ ... ~ + n)a_n
    so this can't work.

    This argument may not quite work either. Someone will have to double check me on this.

    As you suggested, write this as
    \lim_{n \to \infty} \frac{a_1 + 2a_2 + ~ ... ~ + na_n}{1 + 2 + ~ ... ~ + n} = \lim_{n \to \infty} \frac{\sum_{k = 1}^nka_k}{\sum_{k = 1}^n k}

    Now, because (a_n) = A we know that for some n = N (where N is sufficiently large) that a_n is arbitrarily close to A. Similarly for a sufficiently large M (not necessarily the same as N, but we may take it to be as least as large as N) the fraction
    \frac{1}{1 + 2 + ~ ... ~} is arbitrarily close to \frac{1}{M + (M + 1) + ~ ... ~}. Thus:
    \lim_{n \to \infty} \frac{\sum_{k = 1}^nka_k}{\sum_{k = 1}^n k} = \lim_{n \to \infty} \frac{\sum_{k = N}^n kA}{\sum_{k = M}^n k}

    Now pick the larger of N and M and call it P. Then
    \lim_{n \to \infty} \frac{\sum_{k = 1}^nka_k}{\sum_{k = 1}^n k} = \lim_{n \to \infty} \frac{\sum_{k = P}^n kA}{\sum_{k = P}^n k} = A

    I think this will work, but I may have skipped crucial details.

    -Dan
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