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(-1)^n/(n)! find the sum from n=0 to n=infinity

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data

    (-1)^n/(n)! find the sum from n=0 to n=infinity and do the same for (-1)^n/(2n)! Show all work

    2. Relevant equations



    3. The attempt at a solution

    First answer is .367876. I just dont know how to expand out (2n)!. I know (n)! is expanded like: 1*1, 1*2, 1*2*3, 1*2*3*4...... and etc. However I don't know how the expanded version of (2n)! works.
     
  2. jcsd
  3. Sep 27, 2011 #2
    Re: Series

    You have likely derived Taylor Expansions for some common functions, like the exponential and trigonometric functions. Can you use these to evaluate your sum?
     
  4. Sep 27, 2011 #3

    SammyS

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    Re: Series

    It looks like (-1)n/(n)! should give 1/e = e-1 ≈0.36787944117144233

    Write the first several terms for each series. Follow lineintegral1's suggestions

    Also, (2n)! = 1*2*3*4* ... *(n-2*(n-1)*(n)*(n+1)*(n+2)* ... *(2n-2)*(2n-1)*(2n).

    (2*0)! = 0! = 1

    (2*1)! = 2! = 2

    (2*2)! = 4! = 24

    (2*3)! = 6! = 720

    ...
     
    Last edited: Sep 27, 2011
  5. Sep 27, 2011 #4
    Re: Series

    Thank you so much!
     
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