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Math Help - series

  1. #1
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    series

    Hey All,

    Can someone check if I have done this right. Thank you


    find the range of values for x for which:
    ____
    s = 9^nx^{2n} / n^3
    __n=1

    Un + 1 = 9^n.9.x^{2n}.x^2 / n^3 + 3n^2 + 3n + 1

    (Un+1)/(Un)= 9^n.9.x^{2n}.x^2 / n^3 + 3n^2 + 3n + 1 . n^3 / 9^n + x^{2n}

    simplifying

    9x^2.n^3 / n^3 + 3n^2 + 3n + 1

    dividing n^3

    9x^2 / 1 + 3/n + 3n^2 + 1/n^3

    n→1

    9x^2 / 8
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dadon View Post
    Hey All,

    Can someone check if I have done this right. Thank you


    find the range of values for x for which:
    ____
    s = 9^nx^{2n} / n^3
    __n=1

    Un + 1 = 9^n.9.x^{2n}.x^2 / n^3 + 3n^2 + 3n + 1

    (Un+1)/(Un)= 9^n.9.x^{2n}.x^2 / n^3 + 3n^2 + 3n + 1 . n^3 / 9^n + x^{2n}

    simplifying

    9x^2.n^3 / n^3 + 3n^2 + 3n + 1

    dividing n^3

    9x^2 / 1 + 3/n + 3n^2 + 1/n^3

    n→1

    9x^2 / 8
    Not sure how you ended up with 9x^2 / 8, but i suppose you want the values of x for which S converges.

    Here,
    Attached Thumbnails Attached Thumbnails series-series.gif  
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  3. #3
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    how do you know it is less than 1?

    thanks
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  4. #4
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    Quote Originally Posted by dadon View Post
    how do you know it is less than 1?

    thanks
    Because the theorem says that the ratio limit must be strictly less than 1 for the series to converge.
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