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Math Help - Infinite Sums

  1. #1
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    Infinite Sums

    How can I evaluate the sum from i=0 to infinity of i^2 / (4^i) ?

    Also, I would like to check my other answers. For sum i=0-> inf of 1/4^i I got 4/3

    For sum i=0-> inf of i/4^i I got 4/9

    Sorry this looks kind of crappy.
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  2. #2
    Member Traveller's Avatar
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    Your other answers are correct.

    Let S = \frac{1^2}{4} + \frac{2^2}{4^2} + \frac{3^2}{4^3} + ...

    We already know that S'(say) = \frac{1}{4} + \frac{2}{4^2} + \frac{3}{4^3} + ... = \frac{4}{9}

    Observe that S - \frac{S}{4} = \frac{1^2}{4} + ( \frac{2^2}{4^2} - \frac{1^2}{4^2})+ (\frac{3^2}{4^3} - \frac{1^2}{4^2}) + ...

    \Rightarrow \frac{3S}{4} = \frac{1^2}{4} +  \frac{(2+1)(2-1)}{4^2}  + \frac{(3+2)(3-2)}{4^3} + ...

    \Rightarrow \frac{3S}{4} = \frac{1}{4} +  \frac{(2+1)}{4^2}  \frac{(3+2)}{4^3} + ...

    \Rightarrow \frac{3S}{4} = (\frac{1}{4} + \frac{2}{4^2} + \frac{3}{4^3} + ...) + ( \frac{1}{4^2} + \frac{2}{4^3} + \frac{3}{4^4} + ...)

    \Rightarrow \frac{3S}{4} = S' + \frac{S'}{4} = \frac{5S'}{4}<br />
    \Rightarrow S = \frac{20}{27}
    Last edited by Traveller; September 25th 2010 at 11:20 AM.
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  3. #3
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    Thanks for answering, but I don't see how you got your answer for S-S/4.

    Shouldn't the numerators in S/4 be increasing, not just be 1^2? When I subtract S - S/4, I get 1/4 + 1/16 + 1/64 ... so the numerators are always 1. Am I missing something?
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  4. #4
    Member Traveller's Avatar
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    Quote Originally Posted by veronicak5678 View Post
    Thanks for answering, but I don't see how you got your answer for S-S/4.

    Shouldn't the numerators in S/4 be increasing, not just be 1^2? When I subtract S - S/4, I get 1/4 + 1/16 + 1/64 ... so the numerators are always 1. Am I missing something?
    Numerators remain the same, only denominators are multiplied by 4. Our goal was to bring the sum into a form that has i instead of i^2 in the numerator. We achieved this by using the identity a^2 - b^2 = (a+b)(a-b).
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