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Math Help - SUM of N / 2^N

  1. #1
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    SUM of N / 2^N

    Hello:

    I wonder how do you prove that the SUM of N / 2^N (N=1 to infinity) is N / (N-1)^2.

    Thanks!
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  2. #2
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    Quote Originally Posted by maltz View Post
    how do you prove that the SUM of N / 2^N (N=1 to infinity) is N / (N-1)^2.
    One does not because \sum\limits_{N = 1}^\infty  {\frac{N}{{2^N }}}  = \frac{{\left( {\frac{1}{2}} \right)}}{{\left( {1 - \frac{1}{2}} \right)^2 }} = 2.
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  3. #3
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    Ops I am sorry. I wrote the question wrong.
    It should be SUM (N / K^N) (N = 1 to infinity, K is a positive integer),
    and the result is N/(K-1)^2

    What is the formula? Thanks again.
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  4. #4
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    Once again you have written a false statement.
    This is true:
    \sum\limits_{N = 1}^\infty  {\frac{N}{{K^N }}}  = \frac{{\left( {\frac{1}{K}} \right)}}{{\left( {1 - \frac{1}{K}} \right)^2 }} = \frac{K}{{\left( {K - 1} \right)^2 }},\;K > 1.
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  5. #5
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    Hello, maltz!

    Please check the original wording of the problem.
    . . As written, it makes no sense.


    Prove: . \sum^{\infty}_{n=1}\frac{n}{2^n} \;=\;\frac{n}{(n-1)^2}

    If n \to \infty, there would be no n on the right side.


    If you meant: . \sum^n_{k=1}\frac{k}{2^k} \;=\;\frac{n}{(n-1)^2} , .that makes more sense,
    . . . . but unfortunately, it is not true.

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  6. #6
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    Yeah I suck... sorry.
    So it is SUM (N/K^N) = K/(K-1)^2
    I am just wondering how you can proof it instead of memorizing a formula.
    That would be greatly appreciated.

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  7. #7
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    \sum\limits_{N = 0}^\infty  {x^N }  = \frac{1}{{1 - x}},\;\left| x \right| < 1

    Differentiate: \sum\limits_{N = 1}^\infty  {Nx^{N - 1} }  = \frac{1}{{\left( {1 - x} \right)^2 }}.

    Multiply by x: \sum\limits_{N = 1}^\infty  {Nx^N }  = \frac{x}{{\left( {1 - x} \right)^2 }}.

    Let x = \frac{1}{K},\;K > 1
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