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Series Converges to?

  1. Dec 12, 2009 #1
    1. The problem statement, all variables and given/known data
    I have the series [tex]\sum\frac{b^{(2n+2)}(-1)^{n}}{(2n+2)!}[/tex] from n=0 to infinity. I am trying to find what it converges to in terms of b.


    2. Relevant equations
    Using the Ratio Test I have established that it does converge.


    3. The attempt at a solution
    I have scoured the internet, my notes, and all my books, but I can't seem to find a way to find what these kinds of series (power I believe) converge to, only ways to see if they converge or not. I just need to find out the method to calculate what it converges to in terms of b. Thank you for any assistance.
     
  2. jcsd
  3. Dec 12, 2009 #2

    zcd

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    [tex]\sum\frac{b^{(2n+2)}(-1)^{n}}{(2n+2)!}=\sum\frac{b^{(2n+2)}(-1)^{n}}{(2n+2)(2n+1)!}[/tex], from which you can differentiate term by term
     
  4. Dec 12, 2009 #3
    Why would I differentiate it? How does that help find what it converges to?
     
  5. Dec 12, 2009 #4

    zcd

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    [tex]\frac{d}{dx}\sum\frac{b^{(2n+2)}(-1)^{n}}{(2n+2)(2n+1)!}=\sum\frac{b^{(2n+1)}(-1)^{n}}{(2n+1)!}[/tex] which looks a bit like which function?
     
  6. Dec 12, 2009 #5
    It looks like the general term of the Taylor Polynomail for sin(x)...so it is sin(b)?
     
  7. Dec 12, 2009 #6

    zcd

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    The derivative of the series converges to sin(b), not the original series.
     
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