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Sum to infinity question

  1. Jan 22, 2014 #1
    1. The problem statement, all variables and given/known data

    Sum to infinity

    [tex] \frac{1}{2!} - \frac{ \pi ^2}{4^2.4!} + \frac{\pi^4}{4^4.6!} ..... [/tex]
    2. Relevant equations



    3. The attempt at a solution

    I thought the series was similar to the Maclaurin expansion of cos x

    so I tried putting in x= ∏/4

    But I end up with the series [tex] {1} - \frac{ \pi ^2}{4^2.2!} + \frac{\pi^4}{4^4.4!} .....[/tex]

    I don't know how to change the factorials to make it match the ones in the series given.
     
  2. jcsd
  3. Jan 22, 2014 #2

    haruspex

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    Try matching up the terms of the two series based on the factorials, not on the powers of pi. What do you notice?
     
  4. Jan 22, 2014 #3
    Well the factorials increase by 2. so it's like (2n)!?
     
  5. Jan 22, 2014 #4

    Integral

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    You may want to look a bit closer at how you factored 1/2! out.

    4! = 2! *3*4 <> 2*2!
     
  6. Jan 22, 2014 #5
    I know 4! is not equal to 2 x 2!. I didn't factor out anything all I did was put in x = pi/4 in the Maclaurin expansion of CosX
     
  7. Jan 22, 2014 #6

    pasmith

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    You can write this as [itex]F(\pi/4)[/itex], where
    [tex]
    F(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+2)!}x^{2n}
    [/tex]
    so that
    [tex]
    x^2F(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+2)!}x^{2n+2}
    = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2(n+1))!}x^{2(n+1)}
    [/tex]
     
  8. Jan 22, 2014 #7
    Multiply the whole series by (π/4)2 and see what you get!
     
  9. Jan 22, 2014 #8

    haruspex

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    Yes. Compare the term with (2n)! in the first series with the term with (2n)! in the second series, same n.
     
  10. Jan 22, 2014 #9
    [tex]s= \frac{1}{2!} - \frac{ \pi ^2}{4^2.4!} + \frac{\pi^4}{4^4.6!} ..... [/tex]

    [tex](\frac{π}{4})^2s= \frac{(\frac{π}{4})^2}{2!}-\frac{(\frac{π}{4})^4}{4!}+\frac{(\frac{π}{4})^6}{6!} ..... =1-\cos{\frac{π}{4}}[/tex]

    [tex]s=(\frac{4}{π})^2(1-\cos{\frac{π}{4}})=(\frac{4}{π})^2(1-\frac{1}{\sqrt{2}})[/tex]
     
  11. Jan 22, 2014 #10
    Wow Chestermiller, I feel kind of stupid now. Thank you... I was thinking that it looked like cosx but the factorials weren't adding up, I should of tried to get the powers up so it could match. Thank you again guys.
     
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