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Math Help - Series Limit Problem

  1. #1
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    Series Limit Problem

    Consider SN=sum(n=1 to N) (1/(n^2+2n))

    Write a formula for SN and find lim n-->infinity SN


    So the terms are like 1/3+1/8+1/15+1/24.... essentially we are adding 5,7,9,11...etc..the odd numbers starting from 5 to the bottom. I don't know how to write a formula for this that is not recursive though.

    I got SN=(SN-1)/(2N+1), but I think they want a formula that doesn't use recursion.
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  2. #2
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    Quote Originally Posted by zhupolongjoe View Post
    Consider SN=sum(n=1 to N) (1/(n^2+2n))

    Write a formula for SN and find lim n-->infinity SN


    So the terms are like 1/3+1/8+1/15+1/24.... essentially we are adding 5,7,9,11...etc..the odd numbers starting from 5 to the bottom. I don't know how to write a formula for this that is not recursive though.

    I got SN=(SN-1)/(2N+1), but I think they want a formula that doesn't use recursion.
    Note that \frac{1}{n^2+2n} = \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2} \right). Write out some of the terms and see what happens.
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  3. #3
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    Quote Originally Posted by zhupolongjoe View Post
    Consider SN=sum(n=1 to N) (1/(n^2+2n))

    Write a formula for SN and find lim n-->infinity SN


    So the terms are like 1/3+1/8+1/15+1/24.... essentially we are adding 5,7,9,11...etc..the odd numbers starting from 5 to the bottom. I don't know how to write a formula for this that is not recursive though.

    I got SN=(SN-1)/(2N+1), but I think they want a formula that doesn't use recursion.

    Elephant hint: \frac{1}{n(n+2)}=\frac{1}{2}\,\left(\frac{1}{n}-\frac{1}{n+2}\right)

    Tonio
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  4. #4
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    Thanks, is it 3/4?
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  5. #5
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    Quote Originally Posted by zhupolongjoe View Post
    Thanks, is it 3/4?

    Yup

    Tonio
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