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Math Help - series problem

  1. #1
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    series problem

    I am having difficulty solving this math problem:
    Suppose it is known that |x| < 1. What then is the sum from n=1 to n=infinity of nx^n?

    I know how to do it if there is a constant in front of the x, but I don't know what to do since there's a variable in front of it.
    Anybody have any ideas, it's really starting to annoy me?

    Thanks a Lot
    -Nertil
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  2. #2
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    Quote Originally Posted by nertil1
    I am having difficulty solving this math problem:
    Suppose it is known that |x| < 1. What then is the sum from n=1 to n=infinity of nx^n?

    I know how to do it if there is a constant in front of the x, but I don't know what to do since there's a variable in front of it.
    Anybody have any ideas, it's really starting to annoy me?

    Thanks a Lot
    -Nertil
    *Warning*-This is not a formal proof but it captures the idea.

    Let,
    S=x+2x^2+3x^3+4x^4+...
    Thus,
    xS=x^2+2x^3+3x^4+4x^5+...
    Let,
    S'=x+x^2+x^3+x^4+...
    Thus,
    xS+S'=x+2x^2+3x^3+4x^4+...
    Thus,
    xS+S'=S
    Thus,
    \frac{S'}{1-x}=S
    But,
    S'=\frac{x}{1-x} because it is an infinite geometric series,
    Thus,
    S=\frac{x}{(1-x)^2}
    Last edited by ThePerfectHacker; March 6th 2006 at 07:40 PM.
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  3. #3
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    I was thinking about another way of doing this problem.
    Notice the partial sums,
    S_1=x
    S_2=x+2x^2=(x+x^2)+x^2=x\left(\frac{1-x^2}{1-x}\right)+x^2
    S_3=x+2x^2+3x^3=(x+x^2+x^3)+(x^2+x^3)+x^3= x\left(\frac{1-x^3}{1-x}\right)+x^2\left(\frac{1-x^2}{1-x}\right)+x^3
    In general,
    S_n=\frac{1}{1-x}[x(1-x^n)+x^2(1-x^{n-1})+...+x^n]
    All of this is a manipulation of the geometric series,

    As you take the limit the exponents die,
    Thus,
    S_{\infty}=\frac{1}{1-x}[x+x^2+x^3+..]
    But that is the geometric series thus,
    S_{\infty}=\frac{x}{(1-x)^2}

    This proof is more formal.
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  4. #4
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    I realized yet one more way, but this is super informal.
    Integrate part by part,
    x+2x^2+3x^3+4x^4+...
    Thus,
    \frac{x^2}{2}+\frac{2x^3}{3}+\frac{3x^4}{4}+...
    Express as,
    (-x+x)+\left(-\frac{x}{2}+x^2\right)+\left(-\frac{x^2}{3}+x^3\right)+...
    Express as,
    \left(-x-\frac{x^2}{2}-\frac{x^3}{3}-...\right)+(x+x^2+x^3+...)
    Know the expansion for infinite series we have,
    \ln|1-x|+\frac{x}{1-x}
    Thus, the derivative of this must be the initial series
    -\frac{1}{1-x}+\frac{1}{(1-x)^2}=\frac{x}{(1-x)^2}
    ----------------
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  5. #5
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    how did you get an S prime in the first solution?
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  6. #6
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    Quote Originally Posted by nertil1
    how did you get an S prime in the first solution?
    As |x|<1 consider:

    <br />
\frac{1}{1-x}=1+x+x^2+...+x^n+...<br />

    Now differentiate both sides:

    <br />
\frac{1}{(1-x)^2}=1+2x+3x^2+...+nx^{n-1}+...<br />
,

    so multiplying through by x gives:

    <br />
\frac{x}{(1-x)^2}=x+2x^2+3x^3+...+nx^n+...<br />

    You might want to worry about convergence, if you check you should find
    the rules for term by term differentiation are satisfied.

    RonL
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