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Math Help - Simple limit problem

  1. #1
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    Simple limit problem

    hello,
    I'm trying to recall how to do this and I got stuck big time... I would be most grateful for any help

    problem is :

    I'll fix it in latex as soon it work with limits


    \lim_{n \to \infty} ( \frac {1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2} + ... + \frac{n-1}{n^2} )

    and I recall that "limit of sum" is equal to "sum of limits"... something like ::
    if is \lim_{n \to \infty} f(x) = A and \lim_{n \to \infty} f(x) = B than

    \lim_{n\to \infty} ( f(x) +- g(x) ) = lim_{n\to \infty} f(x) +- lim_{n\to \infty} g(x)

    and if I apply that to problem that I have, result would be zero but book says something else so now I'm in big trouble....

    I also have more or less the same another problem but it's to "ugly" to write it without LATEX so...


    what do I do wrong ?
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  2. #2
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    It will probably help if you realise that since there's a common denominator, you can write

    1/n^2 + 2/n^2 + 3/n^2 + ... + (n - 1)/n^2 = [1 + 2 + 3 + ... + (n-1)]/n^2 = [n(n-1)/2]/n^2 = (n^2 - n)/2n^2 = 1/2 - 1/2n.

    I'm sure you can take the limit now.
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  3. #3
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    Thank you very very much

    I'm just curios now, why I can't apply way that I was going...

    and if I have sequence with just odd numbers .... (1+3+5+....+(2n-1) )/(n+1) I would have what ? I think I don't really see from what do u conclude that it's [n(n-1)/2]/n^2 ...
    hehehe or I'm missing basic stuff here
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  4. #4
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    The top is still an arithmetic series, so you can use S = N(t_1 + t_N)/2 to simplify...
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  5. #5
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    ah... OK
    meaning that every time that I have infinite sequence like 1+2+3+... +n I have to look at it like series and go to nth partial sum
    Thanks
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  6. #6
    Senior Member yeKciM's Avatar
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    or simple use Stolz th that says ....

    if you are given two sequences (x_n) and (y_n) and if
    1 lim_(n->\infty) (x_n) = +\infty
    2 y_(n+1) >= y_n

    than if exist

    \lim_{n\to \infty} \frac {x_{n+1}-x_n }{y_{n+1}-y_n }

    and it's equal

    \lim_{n\to \infty} \frac {x_n}{y_n}=\lim_{n\to \infty} \frac {x_{n+1}-x_n }{y_{n+1}-y_n }
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