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Math Help - Partial fraction decomposition

  1. #1
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    Partial fraction decomposition

    I have this  a_n=\frac{1}{n^2+1} and I have to rewrite this to partial fraction decomposition.

    like this?

     \frac{n(x)}{d(x)}= \frac{1}{n}+\frac{1}{n+1}
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    Quote Originally Posted by Anemori View Post
    I have this  a_n=\frac{1}{n^2+1} and I have to rewrite this to partial fraction decomposition.

    like this?

     \frac{n(x)}{d(x)}= \frac{1}{n}+\frac{1}{n+1}
    \frac{1}{n^2+1} is not factorable ...

    why do you need to perform the partial fraction decomposition?
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    Quote Originally Posted by skeeter View Post
    \frac{1}{n^2+1} is not factorable ...

    why do you need to perform the partial fraction decomposition?

    I have this problem:

    Consider a_n=\frac{1}{n^2+1}

    Then,
    a.) Rewrite  a_n using partial decomposition.

    b.) Using a.) to help show the Sum_\infty=1

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    Quote Originally Posted by Anemori View Post
    I have this problem:

    Consider a_n=\frac{1}{n^2+1}

    Then,
    a.) Rewrite  a_n using partial decomposition.

    b.) Using a.) to help show the Sum_\infty=1

    you need to check the problem statement again ...

    \sum_{n=1}^\infty \frac{1}{n^2+1} \ne 1
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    Quote Originally Posted by skeeter View Post
    you need to check the problem statement again ...

    \sum_{n=1}^\infty \frac{1}{n^2+1} \ne 1

    That is what it says in the problem...
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    Quote Originally Posted by Anemori View Post
    That is what it says in the problem...
    the problem statement is wrong.

    this sum is true ...

    \sum_1^{\infty} \frac{1}{n(n+1)} = 1
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    Quote Originally Posted by skeeter View Post
    the problem statement is wrong.

    this sum is true ...

    \sum_1^{\infty} \frac{1}{n(n+1)} = 1
    I see thanks... did you use telescoping series?


    still having problem rewriting it to partial fraction decomposition...
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  8. #8
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    Quote Originally Posted by Anemori View Post
    I see thanks... did you use telescoping series?


    still having problem rewriting it to partial fraction decomposition...
    yes

    \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}

    1  = A(n+1) + Bn

    let n = -1 ... B =  -1

    let n = 0 ... A = 1

    \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}
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  9. #9
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    Quote Originally Posted by skeeter View Post
    yes

    \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}

    1  = A(n+1) + Bn

    let n = -1 ... B =  -1

    let n = 0 ... A = 1

    \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}
    I'm sorry i got wrong equation. but it should be the same... its  a_n=\frac{1}{(n^2+n)}

    i look it up and it was the same... thanks!
    Last edited by Anemori; May 9th 2010 at 03:44 PM. Reason: wrong equation
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  10. #10
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    Quote Originally Posted by Anemori View Post
    I'm sorry i got wrong equation. but it should be the same... its  a_n=\frac{1}{(n^2+n)}

    i look it up and it was the same... thanks!
    ... and you are very welcome.
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