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Math Help - sequences and series

  1. #1
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    Exclamation sequences and series

    Please i don't know how to even BEGIN this question.

    A sequence defined as Sn(x) 1 + 2x + 3x^2.....nx^(n-1)

    (I assume that the it is a geometric sequence and a is n and r is x? Am i correct?)



    By considering (1-x)Sn(x) show that

    Sn(x) = 1-x^n/(1-x) - nx^n/(1-x)^2

    Obtain the value of

    lim (n tends to infinity) {2/3 + 3/2^2 + 4/3^3 + ..... + n/3^(n-1) + 3/2*n/3^n }

    What do they mean by limit and i just dont get it! Please help someone!!!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathshelpneeded View Post
    A sequence defined as Sn(x) 1 + 2x + 3x^2.....nx^(n-1)

    (I assume that the it is a geometric sequence and a is n and r is x? Am i correct?)
    No, because 2x/1 = 2x and 3x^2/(2x) = (3/2)x. The terms are not related by a constant factor.

    Quote Originally Posted by mathshelpneeded View Post
    By considering (1-x)Sn(x) show that

    Sn(x) = 1-x^n/(1-x) - nx^n/(1-x)^2
    Your written answer is incorrect. (And you really need to use parenthesis!)

    Sn(x) = 1 + 2x + 3x^2 + ... + nx^{n - 1}

    Now
    1*Sn(x) = 1 + 2x + 3x^2 + ... + nx^{n - 1}


    x*Sn(x) = x + 2x^2 + 3x^3 + ... + nx^n

    So
    (1 - x)*Sn(x) = 1 + x + x^2 + ... + x^{n - 1} - nx^n

    Now 1 + x + x^2 + ... + x^{n - 1} is a geometric series with n terms, so the sum of this is:
    (1 - x^n)/(1 - x)

    Thus:
    (1 - x)*Sn(x) = (1 - x^n)/(1 - x) - nx^n

    Sn(x) = (1 - x^n)/(1- x)^2 - nx^n/(1 - x)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathshelpneeded View Post
    Obtain the value of

    lim (n tends to infinity) {2/3 + 3/2^2 + 4/3^3 + ..... + n/3^(n-1) + 3/2*n/3^n }

    What do they mean by limit and i just dont get it! Please help someone!!!
    In this case, by "limit" they mean they want the exact value of the series if the series has an infinite number of terms.


    Note that
    2/3 + 3/3^2 + 4/3^3 + ...
    is just your Sn(x) series for x = 1/3 with 1 subtracted, or:
    2/3 + 3/3^2 + 4/3^3 + ... = Sn(1/3) - 1 (for n -> infinity)

    So
    Sn(1/3) = [1 - (1/3)^n]/[1 - (1/3)]^2 - n*(1/3)^n/[1 - (1/3)]

    = (9/4)*[1 - 1/3^n] - (3/2)*n/3^n

    Now, for very large n, what happens to the first term? I'll let you work it on a calculator and simply say that as n grows very large, 3^n grows very large, and thus 1/3^n goes to 0. So the first term goes to a value of 9/4 as n goes to infinity.

    For very large n, what happens to the second term? Using a calculator you can again verify that this term also goes to 0.

    Thus as n goes to infinity:
    Sn(x) -> 9/4.

    Finally:
    2/3 + 3/3^2 + 4/3^3 + ... = Sn(1/3) - 1 (for n -> infinity)

    = 9/4 - 1 = 5/4.

    -Dan
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  4. #4
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    Quote Originally Posted by topsquark View Post

    Note that
    2/3 + 3/3^2 + 4/3^3 + ...
    is just your Sn(x) series for x = 1/3 with 1 subtracted, or:
    Thanks a lot i never noticed that.
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