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Find the Sum of this Alternating Series

  1. Oct 29, 2014 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Find the sum of
    starts at 0 to infinity ∑ (cos(k*pi))/pi^k

    First, I determined that it does, indeed, converge with the alternating series test.
    Second, I found the answer to be pi/(1+pi) via wolfram alpha.

    But I am at a loss on how to find the answer here.

    This is a geometric series, so I am trying to find the r value. If I can do that, then I can use the trick of first term/(1-r)

    Correct?
     
  2. jcsd
  3. Oct 29, 2014 #2

    Mark44

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    It might be helpful to write the series without the cosine factor. All it contributes is +/- 1 each time.
     
  4. Oct 29, 2014 #3

    LCKurtz

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    Why don't you just divide the ##k+1## term by the ##k## term to get ##r##?
     
  5. Oct 29, 2014 #4

    RJLiberator

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    So, with cos(k*pi) out, that would leave 1/pi^k. Therefore 1/pi^k = r? Hm.

    k+1 term? Ok, let me try what I think you are saying:

    k+1 term = value of 1, so k=1 and therefore cos(1pi)/pi^1 = -1/pi and the original k =0 term = cos(0pi)/pi^0 = 1. So, (-1/pi)/1 = -1/pi.

    Hm, so if r = -1/pi then we can plug that into the first term/(1-r) equation, lets see:
    First term = 1
    (1-(-1/pi) = 1+1/pi
    1/(1+1/pi)

    If you multiple everything by pi, to simplify you get the answer of pi/pi+1 which is the correct answer.

    Why did this work? Is there a trick I am not thinking of?

    Thank you all. It seems I may have made a mechanical error with Mark44's suggestion. LcKurtz, do you know the definition/term of what you suggested so i can research it?
     
  6. Oct 29, 2014 #5

    Mark44

    Staff: Mentor

    No it's not. ##\frac{\pi}{\pi} + 1 = 2##.

    I'm being picky, but you need parentheses like you used in your first post here. IOW, like this: π/(π + 1).
     
  7. Oct 29, 2014 #6

    RUber

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    ## \cos k \pi =(-1)^k ## so you would have ##\sum_k \left(\frac{-1}{\pi}\right)^k ##
    Mark44's suggestion should have led you to the same conclusion for r.
     
  8. Oct 29, 2014 #7

    LCKurtz

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    A geometric series looks like $$
    \sum_{k=0}^\infty ar^k$$where ##r## is the common ratio. The ##k+1## term is ##ar^{k+1}## and the ##k## term is ##ar^k##. When you divide them you get the common ratio ##r##, no matter what ##k## is. What you did is divide the ##k=1## term by the ##k=0## term. That doesn't tell you anything unless you already know it is a geometric series. What you did gave you the correct ratio because it is in fact a geometric series, but the correct way to get the ratio and show it is a geometric series is divide the ##k+1## term by the ##k## term and observe you always get the same answer.
     
  9. Oct 30, 2014 #8

    RJLiberator

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    Excellent analysis, friends. I appreciate the words. Quite frankly, I am getting addicted to this series math.
     
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