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Math Help - [SOLVED] Infinite Sum

  1. #1
    Member Haven's Avatar
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    [SOLVED] Infinite Sum

    Prove that:

    \sum_{i=4}^{\infty} \frac{2}{i(i-1)(i-2)(i-3)} = \frac{1}{3}

    I think I'm supposed to use the fact that any decreasing sequence a_1 \dots a_n can be written as:
     a_n-a_1 = (a_n - a_{n+1}) + \dots (a_3 - a_2) + (a_2 - a_1)
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Haven View Post
    Prove that:

    \sum_{i=4}^{\infty} \frac{2}{i(i-1)(i-2)(i-3)} = \frac{1}{3}

    I think I'm supposed to use the fact that any decreasing sequence a_1 \dots a_n can be written as:
     a_n-a_1 = (a_n - a_{n+1}) + \dots (a_3 - a_2) + (a_2 - a_1)
    Well, something like that, only a little more complicated I think. To get started, you can use a "partial fraction decomposition" like this:

    \sum_{i=4}^\infty \frac{2}{i(i-1)(i-2)(i-3)}=\sum_{i=4}^\infty\left({\color{blue}\frac{1}{3  }\cdot\frac{1}{i-3}-\frac{1}{3}\cdot\frac{1}{i}}+{\color{red}\frac{1}{  i-1}-\frac{1}{i-2}}\right)
    Now determine the sum of the blue and red parts separately.
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  3. #3
    Member Haven's Avatar
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    EDIT: nevermind i figured it out
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Haven View Post
    Prove that:

    \sum_{i=4}^{\infty} \frac{2}{i(i-1)(i-2)(i-3)} = \frac{1}{3}

    I think I'm supposed to use the fact that any decreasing sequence a_1 \dots a_n can be written as:
     a_n-a_1 = (a_n - a_{n+1}) + \dots (a_3 - a_2) + (a_2 - a_1)
    There is a much more general way to solve these kind. I.e. \sum_{n=0}^{\infty}\frac{1}{(n+1)\cdots(n+\ell)}=\  frac{1}{\ell(\ell-1)(\ell-1)!}. Ask to see a proof if you're interested.
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