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Math Help - Prove that series = ... ?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove that series = ... ?

    Prove that:

    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 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: S_n=\sum_{k=0}^{\infty}{\tfrac{\binom{n^2}{k}}{\bi  nom{n+k}{k}}\cdot (-1)^k}<br />

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

    So: <br />
\tfrac{1}<br />
{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 } <br />
=<br />
\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: <br />
\tfrac{1}<br />
{n} \cdot S_n  = \int_0^1 {\left( {1 - x} \right)^{n^2 }  \cdot \left( {1 - x} \right)^{n - 1} dx}  = \tfrac{1}<br />
{{n \cdot \left( {n + 1} \right)}}<br />
and we are done \square

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