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Math Help - Series problem with limiting sum

  1. #1
    Member jacs's Avatar
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    Series problem with limiting sum

    A limit problem i have tried for hours now and just cannot wrap my head around
    Solution is apparently (1 + x)/(1 - x)^2

    any help greatly appreciated,
    thanks
    jacs
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  2. #2
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    We are adding all of these together. So let us add them in a convenient way.

    First we add the left-most coloums together:
    1+x+x^2+x^3+... = \frac{1}{1-x}

    Second we add the next left-most coloums together:
    2x+2x^2+2x^3+... = 2x(1+x+x^2+...) = \frac{2x}{1-x}

    Third we add the next left-most colums together:
    2x^2+2x^3+2x^4+... = 2x^2(1+x+x^2+...) = \frac{2x^2}{1-x}

    And so on....

    So in total we need to add,
    \frac{1}{1-x} + \frac{2x}{1-x}+\frac{2x^2}{1-x}+\frac{2x^3}{1-x}+...

    I am going to add and subtract \frac{1}{1-x} to make it an easier form.

    \left(\frac{2}{1-x} + \frac{2x}{1-x}+\frac{2x^2}{1-x}+\frac{2x^3}{1-x}+... \right) - \frac{1}{1-x}
    Factor again,
    \frac{2}{1-x}\left( 1+x+x^2+x^3+... \right) - \frac{1}{1-x}

    \frac{2}{(1-x)^2} - \frac{1}{1-x} =\boxed{ \frac{1+x}{(1-x)^2}}
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  3. #3
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    Hello, jacs!


    Observe that:
    . . 1 \:=\:1
    . 3x\:=\:x + 2x
    5x^2\:=\:x^2 + 2x^2 + 2x^2
    7x^3 \:=\:x^3 + 2x^3 + 2x^3 + 2x^3
    9x^4 \:=\:x^4 + 2x^4 + 2x^4 + 2x^4 + 2x^4

    By studying the arrangement, or otherwise, find in simplest algebraic form,
    . . an expression for the limit of the series: . 1 + 3x + 5x^2 + 7x^3 + 9x^4 + \cdots

    Solution: . \frac{1+x}{(1-x)^2}
    Add the stack of equations . . .

    The left side is: . S\:=\:1 + 3x + 5x^2 + 7x^3 + 9x^4 + \cdots

    The right side is: . (1 + x + x^2 + \cdots) + (2x + 2x^2 + 2x^3 + \cdots) + (2x^2 + 2x^3 + 2x^4+\cdots) + \cdots

    . . . . . . . . . . . = \;(1 + x + x^2 + \cdots) + 2x(1 + x + x^2 + \cdots) + 2x^2(1 + x + x^2 + \cdots) + \cdots

    Factor: . (1 + x + x^2 + \cdots)\cdot[1 + 2x + 2x^2 + 2x^3 + \cdots]

    . . . . = \;(1 + x + x^2 + \cdots)\cdot\left[1 + 2x(1 + x + x^2 + \cdots)\right]

    The expressions in parentheses are geometric series:
    . . first term a = 1, common ratio r = x
    Their sum is: . \frac{1}{1-x}

    Hence, we have: . S \;=\;\frac{1}{1-x}\cdot\left[1 + 2x\left(\frac{1}{1-x}\right)\right]

    . . which simplifies to: . \boxed{S \;=\;\frac{1+x}{(1-x)^2}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Another solution . . . (They did say "otherwise", right?)


    We are given: . S \;=\;1 + 3x + 5x^2 + 7x^3 + 9x^4 + \cdots

    Multiply by x\!:\;xS \;=\;\qquad x + 3x^2 + 5x^3 + 7x^4 + \cdots

    Subtract: . S - xS \;=\;1 + 2x + 2x^2 + 2x^3 + 2x^4 + \cdots

    . . . . . . . (1 - x)S \;=\; 1 + 2x\underbrace{(1 + x + x^2 + x^3 +\cdots)}_{\text{geometric series}}

    So we have : . (1 - x)S \;=\;1 + 2x\left(\frac{1}{1-x}\right) \;=\;\frac{1+x}{1-x}

    Therefore: . S \;=\;\frac{1+x}{(1-x)^2}

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  4. #4
    Member jacs's Avatar
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    thanks to both of you. i dont think i ever woudl have figured that one out in a milion years. Going ot have a study of all the different methods and see which one i can try to replicate without giving myself and annureism.

    thanks, you guys rock!
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