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Evaluating an infinite series

  1. Mar 1, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]
    \sum_{n=1}^{\infty}\frac{sinn}{2^n}
    [/tex]


    2. Relevant equations
    Definition of a geometric series:
    [tex]
    \sum_{n=0}^{\infty}x^n=\frac{1}{1-x}
    [/tex]



    3. The attempt at a solution
    Basically I can use the geometric series idea and implement it into the denominator of the question (i.e. sub x=2 into the equation from part b and change the lower index to n=1)
    [tex]
    \sum_{n=1}^{\infty}2^n=\frac{1}{1-2}
    [/tex]

    Taking the derivate of both sides:
    [tex]
    \sum_{n=1}^{\infty}n2^{n-1}=\frac{1}{(1-2)^2}
    [/tex]

    Multiplying both sides by 2:
    [tex]
    \sum_{n=1}^{\infty}n2^n=\frac{2}{(1-2)^2}
    [/tex]

    Of course I can simplify the question furthermore and get a single value as a result, my problem is I don't understand on how to incorporate the numerator (sin n) into the problem. Is my approach correct, or am I substituting the wrong x-value?

    Thanks in advance for any help/advice. :)
     
  2. jcsd
  3. Mar 1, 2010 #2

    Mark44

    Staff: Mentor

    What exactly are you supposed to do with this series?
     
  4. Mar 1, 2010 #3
    Suppose to solve for the overall sum I'm assuming, the question simply states "Find [tex]

    \sum_{n=1}^{\infty}\frac{sinn}{2^n}

    [/tex]."
     
  5. Mar 1, 2010 #4
    There are two problems here:

    (1) The equation you gave for the geometric series only works for |x| < 1.

    (2) We can't simply change the lower index to n=1.
     
  6. Mar 1, 2010 #5
    Well that was the only approach I could think of, any other method to solve this question? A starting tip/hint will suffice.
     
  7. Mar 1, 2010 #6
    I'm also having some difficulty with the sine in the numerator - I haven't done something like this in a while.

    What class is this for, by the way? (to give me an idea of methods you have)

    -----

    Also, if there is a solution with geometric series, the problems I pointed out can be fixed... you just need to be careful about what you use.
     
  8. Mar 1, 2010 #7
    Calculus II (First year)
     
  9. Mar 1, 2010 #8
    From what I can tell, you can change the lower index to n=1, but I assume I will have to check for convergence first, if the series does converge, I can move ahead with this method, if not, I will need an alternative.
     
  10. Mar 1, 2010 #9
  11. Mar 1, 2010 #10
    I believe we haven't learned that yet unfortunately, however I have determined that the series is absolutely convergent via the comparison test. What r value can i use so [tex]
    |r| < 1
    [/tex], involved with the numerator as well? :S
     
  12. Mar 1, 2010 #11
    Hrmmm... I don't know a solution if you can't use Euler's formula. Sorry.

    (doesn't mean there isn't one, but I'm not seeing it)
     
  13. Mar 1, 2010 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I'm with Mathnerdmo. It's pretty easy if you put r=exp(i)/2 and realize your sum is related to the imaginary part of sum(r^n). So |r|<1 and you can treat it as a geometric series. Are you sure you didn't do that in Calc I and just forgot it?
     
  14. Mar 2, 2010 #13
    After skimming through some info on Euler's formula, I can substitute sinn= (e^in - e^(-in)) / 2i, I'm confused as to what you are trying to state, could you perhaps elaborate? How exactly did you derive with r = exp(i)/2?
     
  15. Mar 2, 2010 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    exp(i)^n=exp(in)=cos(n)+i*sin(n). If you sum that series and take the imaginary part you get the sum of sin(n). What happens if you do the same thing with (exp(i)/2)^n?
     
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