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Find Sum of Series Using Power Series

  1. Jun 28, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the sum of sigma (n=1 to infinity) (-1)^(n-1) * [n/(4^(n-1))] using the power series 1/(1+x) = sigma (n=0 to infinity) (-1)^n * x^n.


    2. Relevant equations
    1/(1+x) = 1 - x + x^2 - x^3 + x^4 + ... + (-1)^n * x^n


    3. The attempt at a solution

    The problem suggested to differentiate both sides and then substitute 1/4 for x.

    I am totally confused and don't even know where to begin.
     
  2. jcsd
  3. Jun 28, 2011 #2

    hunt_mat

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    You know how to differentiate right:
    [tex]
    \frac{d}{dx}\frac{1}{1+x}=-\frac{1}{(1+x)^{2}}
    [/tex]
    I am sure you know how to differentiate [itex]x^{n}[/itex], what do you get if you do this?
     
  4. Jun 28, 2011 #3
    d/dx (x^n) = n * x^(n-1)
     
  5. Jun 28, 2011 #4

    hunt_mat

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    Right, so if you differentiate the series term by term (which you're allowed to do as long as you're within the radius of convergence), what do you get?
     
  6. Jun 28, 2011 #5
    I'm not sure I understand the question.

    Which series?

    Maybe I'm just not understanding the original question.

    1/(1+x) can be rewritten as sum (n=0 to infinity) of (-1)^n * x^n.

    I believe the question is asking me to take the sum (n=1 to infinity) of (-1)^(n-1) * n/4^(n-1) and come up with a small simple equation to determine its sum using the pattern of 1/1+x as a template. Yes?


    I know that if I take x^n and differentiate it I get n/x^n-1, which looks exactly like my n/4^(n-1). Does that mean that the sum of my series is nothing more than 1 - 1/4 + 1/16 - 1/64? Do I need to make an adjustment to my n values because my sum starts @1 and the original starts @0?
     
  7. Jun 28, 2011 #6

    hunt_mat

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    Differentiating the series:
    [tex]
    \sum_{n=0}^{\infty}(-1)^{n}x^{n}
    [/tex]
    is
    [tex]
    \sum_{n=1}^{\infty}(-1)^{n}nx^{n-1}
    [/tex]
    Now what is the above equal to? (hint, we have derived it before)
     
  8. Jun 28, 2011 #7

    vela

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    Do exactly what was suggested. What do you get if you differentiate the lefthand side? What do you get if you differentiate the righthand side? Then plug in x=1/4. What do you get? How is it related to the original question?
     
  9. Jun 28, 2011 #8
    Since I differentiated the right side, I have to differentiate the left side, neh?

    So that will equal -1/(1+x)^2?

    I'm trying not to feel really stupid, and I'm just so frustrated. I greatly appreciate all your help.

    EDIT: When you differentiate do you change the starting value for n? Since I'm not differentiating n (-1)^n stays (-1)^n, shouldn't it?

    EDIT EDIT: Sum = - 16/25?
     
    Last edited: Jun 28, 2011
  10. Jun 28, 2011 #9

    hunt_mat

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    Correct, now following your suggestion, set x=1/4, what will the series look like?
     
  11. Jun 28, 2011 #10
    This is my differentiation:
    [tex]
    \sum_{n=1}^{\infty}(-1)^{n}nx^{n-1}
    [/tex]

    The series was listed as:
    [tex]
    \sum_{n=1}^{\infty}(-1)^{n-1}nx^{n-1}
    [/tex]

    Does the exponent on the (-1) change when differentiated?

    But to answer your question, the series should be:

    1 -1/2 + 3/16 - 1/16 + ...
     
  12. Jun 28, 2011 #11

    hunt_mat

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    The idea here us to note that [itex](-1)^{n-1}=(-1)^{n+1}[/itex] and you have a minus coming from the LHS.
     
  13. Jun 28, 2011 #12
    OH! I see that! If I divide (-1)^n by -1, I get (-1)^n-1. DUH!

    That is how it comes up with being 1/(1+x)^2.

    That makes sense.
     
  14. Jun 28, 2011 #13

    hunt_mat

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    So, now you can tell us what the sum of the series is?
     
  15. Jun 28, 2011 #14
    16/25!!
     
  16. Jun 28, 2011 #15

    hunt_mat

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    Well done!!!
     
  17. Jun 28, 2011 #16
    Thank you SO much for your patience and help. Not only did I understand what I did I can do it on other problems!
     
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