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Math Help - Evaluate the series

  1. #1
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    Evaluate the series

    1. sum of (n/2^n) from n=1 to infinite
    2. sum of (n(n-1)x^n) from n=2 to infinite
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  2. #2
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    Hello, Jojo123!

    Here's the first one . . .


    1)\;\;\sum^{\infty}_{n=1} \frac{n}{2^n}

    We are given: . S \;=\;\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \frac{5}{2^5} +\hdots

    Multiply by \tfrac{1}{2}\!:\;\;\frac{1}{2}S \;=\;\qquad \frac{1}{2^2} + \frac{2}{2^3} + \frac{3}{2^4} + \frac{4}{2^5} + \hdots


    \text{Subtract: }\qquad\quad\frac{1}{2}S \;=\;\underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots}_{\text{geometric series}}

    The geometric series has a sum of 1.

    . . So we have: . \frac{1}{2}S \:=\:1 \quad\Rightarrow\quad\boxed{ S \:=\:2}

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  3. #3
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    Quote Originally Posted by Jojo123 View Post
    2. sum of (n(n-1)x^n) from n=2 to infinite
    f(x)=\sum_{n=0}^\infty x^n=\frac{1}{1-x} (assuming |x|<1)

    hence f'(x)=\sum_{n=1}^\infty nx^{n-1}=\left(\frac{1}{1-x}\right)'=\frac{1}{(1-x)^2}

    f''(x)=\sum_{n=2}^\infty n(n-1)x^{n-2}=\left(\frac{1}{(1-x)^2}\right)'=\frac{1}{2(1-x)^3}
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  4. #4
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    Quote Originally Posted by Jojo123 View Post

    2. sum of (n(n-1)x^n) from n=2 to infinite
    Put \alpha=\sum\limits_{n=1}^{\infty }{nx^{n}}\implies \alpha \cdot x=\sum\limits_{n=1}^{\infty }{nx^{n+1}}. From here it's not hard to see that \alpha =\frac{x}{(1-x)^{2}}. Now put \beta=\sum\limits_{n=1}^{\infty }{n(n-1)x^{n}}\implies \beta \cdot x=\sum\limits_{n=1}^{\infty }{n(n-1)x^{n+1}} and compute \beta-\beta\cdot x.
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