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Math Help - How can I proove that series: (-1)^n (1/n!) = 1/e

  1. #1
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    How can I proove that series: (-1)^n (1/n!) = 1/e

    Hello,

    in case if this is not to difficult I would like some help on prooving the sum of this serie.
    \sum_{n=0}^{fin}(-1)^n\ast {1}/{n!} = 1/e

    Thank you in advance!
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  2. #2
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    Re: How can I proove that series: (-1)^n (1/n!) = 1/e

    Quote Originally Posted by Melsi View Post
    in case if this is not to difficult I would like some help on prooving the sum of this serie.
    \sum_{n=0}^{fin}(-1)^n\ast {1}/{n!} = 1/e
    I find the question a bit strange.
    Are you aware that \left( {\forall x} \right)\left[ {e^x  = 1 + \sum\limits_{k = 1}^\infty {\frac{{x^k }}{{k!}}} } \right]~?

    So in your posting x=~?
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  3. #3
    Grand Panjandrum
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    Re: How can I proove that series: (-1)^n (1/n!) = 1/e

    Quote Originally Posted by Melsi View Post
    Hello,

    in case if this is not to difficult I would like some help on prooving the sum of this serie.
    \sum_{n=0}^{fin}(-1)^n\ast {1}/{n!} = 1/e

    Thank you in advance!
    What do you know about the exponential function and "e"?

    CB
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