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Find the Sum:

  1. Apr 6, 2010 #1
    Find the sum of the series

    E (-1)^(n-1) * (n/(2^{n-1}))

    By using the power series 1/ (1+x) = E (-1)^n *x^n,

    I am unsure of what to do...

    Do I take the derivative with respect to x of both sides?
  2. jcsd
  3. Apr 6, 2010 #2
    now what i did was this:

    1/ (1+x) = E (-1)^n *x^n

    now i took the derivative with respect to x:

    -(1+x)^-2 = E (-1)^n * n*x^(n-1)

    Then i substituted x by 1/2 getting:

    -(1+(1/2))^-2 = E (-1)^(n-1) * (n/ 2^(n-1))

    Does my math look correct? now my sum should be pretty much what i get from this equation:
    but with the substituted 1/2 for x.. right?
  4. Apr 6, 2010 #3


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    You will need to differentiate but lets rewrite your original problem to something more suggestive.

    You can write your original sum as:

    [tex]-2 \sum_n (-1)^n n \left(\frac{x}{2}\right)^n[/tex]

    Compare this to the given power series.
  5. Apr 6, 2010 #4


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    Your maths is almost correct. You make a minus sign error going from step 1 to two. Going from (-1)^n to (-1)^(n-1) requires a multiplication by -1 on both sides.
  6. Apr 6, 2010 #5
    oh ok. so my final answer should be:

    (1+(1/2))^-2 = E (-1)^(n-1) * (n/ 2^(n-1))

    and to find the sum would be to just do the math for:
    (1+(1/2))^-2 = .444
  7. Apr 6, 2010 #6


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    Yes although I would write 4/9 instead.
  8. Apr 6, 2010 #7
    thank you so much!
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