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Thread: Prove that series = ... ?

  1. #1
    Super Member fardeen_gen's Avatar
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    Jun 2008
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    539

    Prove that series = ... ?

    Prove that:

    $\displaystyle 1 - \frac{n^2}{n + 1} + \frac{n^2(n^2 - 1)}{(n + 1)(n + 2)} - \frac{n^2(n^2 - 1)(n^2 - 2)}{(n + 1)(n + 2)(n + 3)} + \mbox{...}\ = \frac{1}{n + 1}$

    where $\displaystyle n\in \mathbb{N}$
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  2. #2
    Super Member PaulRS's Avatar
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    Note that you may re-write the sum as: $\displaystyle S_n=\sum_{k=0}^{\infty}{\tfrac{\binom{n^2}{k}}{\bi nom{n+k}{k}}\cdot (-1)^k}
    $

    Now: $\displaystyle
    n \cdot \int_0^1 {x^k \cdot \left( {1 - x} \right)^{n - 1} dx} =\tfrac{1}{\binom{n+k}{k}}
    $ (you should recognise the Beta function here)

    So:$\displaystyle
    \tfrac{1}
    {n} \cdot S_n = \sum\limits_{k = 0}^\infty {\binom{n^2}{k} \cdot \left( {\int_0^1 {x^k \cdot \left( {1 - x} \right)^{n - 1} dx} } \right) \cdot \left( { - 1} \right)^k }
    $$\displaystyle =
    \int_0^1 {\left( {\sum\limits_{k = 0}^\infty {\binom{n^2}{k} \cdot x^k \cdot \left( { - 1} \right)^k } } \right) \cdot \left( {1 - x} \right)^{n - 1} dx}$

    Now by the Binomial Theorem: $\displaystyle
    \tfrac{1}
    {n} \cdot S_n = \int_0^1 {\left( {1 - x} \right)^{n^2 } \cdot \left( {1 - x} \right)^{n - 1} dx} = \tfrac{1}
    {{n \cdot \left( {n + 1} \right)}}
    $ and we are done $\displaystyle \square$

    Note that you may generalise this, consider $\displaystyle m\in\mathbb{N}$ then: $\displaystyle \sum_{k=0}^{\infty}{\tfrac{\binom{m}{k}}{\binom{n+ k}{k}}\cdot (-1)^k}=
    \tfrac{n}
    {{m + n}}
    $ for all $\displaystyle n\in{\mathbb{Z}^+}$
    Last edited by PaulRS; May 21st 2009 at 06:30 PM.
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