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Thread: pre calculus involving limits only

  1. #1
    Member grgrsanjay's Avatar
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    pre calculus involving limits only

    let $\displaystyle \alpha = \displaystyle\lim_{n\to\infty} \left(\frac{1^2 + 2^2 + ..... +n^2}{n^3}\right) $ and
    $\displaystyle \beta = \displaystyle\lim_{n\to\infty} \left(\frac{(1^3 - 1^2 )+(2^3 -2^2 )+ ..... +(n^3 - n^2 )}{n^4}\right)$ ,then which of the following is correct
    (A)$\displaystyle \alpha = \beta$
    (B)$\displaystyle \alpha < \beta$
    (C)$\displaystyle 4\alpha - 3\beta =0$
    (D)$\displaystyle 3\alpha - 4\beta =0$
    Last edited by grgrsanjay; Aug 31st 2010 at 06:30 AM.
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  2. #2
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    Quote Originally Posted by grgrsanjay View Post
    let $\displaystyle \alpha = \displaystyle\lim_{n\to\infty} \left(\frac{1^2 + 2^2 + ..... +n^2}{n^3}\right) $ and
    $\displaystyle \beta = \displaystyle\lim_{n\to\infty} \left(\frac{(1^3 - 1^2 )+(2^3 -2^2 )+ ..... +(n^3 - n^2 )}{n^4}\right)$ ,then which of the following is correct
    (A)$\displaystyle \alpha = \beta$
    (B)$\displaystyle \alpha < \beta$
    (C)$\displaystyle 4\alpha - 3\beta =0$
    (D)$\displaystyle 3\alpha - 4\beta =0$
    $\displaystyle \alpha$ contains a "sum of squares" in the numerator.

    $\displaystyle \beta$ contains the difference between a "sum of cubes" and a "sum of squares" in the numerator.

    When you express those sums in "closed form", you can evaluate the limits.

    You will find the correct answer is one of the 4 listed.
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  3. #3
    Member grgrsanjay's Avatar
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    Quote Originally Posted by Soroban
    Hello, grgrsanjay!

    Archie Meade gave you an excellent game plan . . . Could you follow it?


    We are expected to know these summation formulas:

    . . $\displaystyle 1^2 + 2^2 + 3^2 + \hdots + n^2 \:=\:\dfrac{n(n+1)(2n+1)}{6}$

    . . $\displaystyle 1^3 + 2^3 + 3^3 + \hdots + n^3 \:=\:\dfrac{n^2(n+1)^2}{2}$


    $\displaystyle \displaystyle \text{Let }\:\alpha \:=\:\lim_{n\to\infty} \left(\frac{1^2 + 2^2 + ..... +n^2}{n^3}\right) \;\;[1]$

    $\displaystyle \displaystyle \text{Let }\:\beta \:=\: \lim_{n\to\infty} \left(\frac{(1^3 - 1^2 )+(2^3 -2^2 )+ \hdots +(n^3 - n^2 )}{n^4}\right) \;\;[2]$


    $\displaystyle \text{Then which of the following is correct?}$

    . . $\displaystyle (A)\; \alpha \:=\: \beta\qquad (B)\;\alpha \:<\: \beta \qquad (C)\; 4\alpha - 3\beta \:=\:0 \qquad (D)\;3\alpha - 4\beta =0$


    In [1], we have: .$\displaystyle \dfrac{1^2+2^2+3^2+\hdots+n^2}{n^3} \;=\;\dfrac{\frac{n(n+1)(2n+1)}{6}}{n^3} \;=\;\dfrac{2n^3+3n^2+n^2}{6n^3}\;=\;\dfrac{1}{3} + \dfrac{1}{2n} + \dfrac{1}{6n^2}$

    . . Hence: .$\displaystyle \displaystyle \alpha \;=\;\lim_{n\to\infty}\left(\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}\right) \;=\;\frac{1}{3}$



    In [2], we have: .$\displaystyle \dfrac{\frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}}{n^4} \;=\;\dfrac{3n^4 + 2n^3 - 3n^2 - 2n}{12n^4} \;=\;\dfrac{1}{4} + \dfrac{1}{6n} - \dfrac{1}{4n^2} - \dfrac{1}{6n^3} $


    Hence: .$\displaystyle \displaystyle \beta \;=\;\lim_{n\to\infty}\left(\frac{1}{4} + \frac{1}{6n} - \frac{1}{4n^2} - \frac{1}{6n^3}\right) \;=\;\frac{1}{4}$


    The correct answer is $\displaystyle (D).$


    yea i could follow it
    it was very easy but i dint use the logic properly
    thank you sorobon
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