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Math Help - Find the sum of the series

  1. #1
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    Question Find the sum of the series

    Find the sum of the series,

    \sum^{\infty}_{n=1} \left( \frac{n+2}{n!} \right)x^{n+1}

    See figure attached for my attempt.

    EDIT: I tried taking another route in my work, see 2nd figure attached for that attempt. (Follow the arrow I added in)
    Attached Thumbnails Attached Thumbnails Find the sum of the series-2009sq.jpg   Find the sum of the series-2009sq2.jpg  
    Last edited by jegues; October 15th 2010 at 01:51 PM.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by jegues View Post
    Find the sum of the series,

    \sum^{\infty}_{n=1} \left( \frac{n+2}{n!} \right)x^{n+1}

    See figure attached for my attempt.
    Are you sure you want the sum and not something else (like the interval of convergence)?
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Are you sure you want the sum and not something else (like the interval of convergence)?
    The only thing the question states is Find the sum of the series. I've added another attempt in the 2nd figure of my OP.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by jegues View Post
    Find the sum of the series,

    \sum^{\infty}_{n=1} \left( \frac{n+2}{n!} \right)x^{n+1}

    See figure attached for my attempt.

    EDIT: I tried taking another route in my work, see 2nd figure attached for that attempt. (Follow the arrow I added in)
    Let consider the series...

    \displaystyle \sum_{n=1}^{\infty} \frac{x^{n+2}}{n!} = x^{2}\ \sum_{n=1}^{\infty} \frac{x^{n}}{n!} = x^{2}\ (e^{x}-1)= f(x) (1)

    Very well!... from (1) You obtain directly...

    \displaystyle \sum_{n=1}^{\infty} \frac{(n+2)}{n!}\ x^{n+1} = f^{'} (x) = x\ (x+2)\ e^{x} -2 x (2)

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by chisigma View Post
    Let consider the series...

    \displaystyle \sum_{n=1}^{\infty} \frac{x^{n+2}}{n!} = x^{2}\ \sum_{n=1}^{\infty} \frac{x^{n}}{n!} = x^{2}\ (e^{x}-1)= f(x) (1)

    Very well!... from (1) You obtain directly...

    \displaystyle \sum_{n=1}^{\infty} \frac{(n+2)}{n!}\ x^{n+1} = f^{'} (x) = x\ (x+2)\ e^{x} -2 x (2)
    Kind regards

    \chi \sigma

    Aha! So my 2nd attempt was correct!

    Thanks again!
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