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Tricky Infinite Sum

  1. Dec 19, 2009 #1
    1. The problem statement, all variables and given/known data

    What is [tex]\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}[/tex]?

    2. Relevant equations
    [tex]e^x[/tex] = [tex]\sum_{n=0}^{\infty}{\frac{x^n}{n!}}[/tex]

    3. The attempt at a solution
    I understand that the given series looks like the series for e^1 but I know that isn't the correct answer. The answer (without solution) was supplied as (1/2)(e+e^-1). I can't seem to figure out the extra e^-1 part. Help?
  2. jcsd
  3. Dec 19, 2009 #2


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    Science Advisor

    Try writing out e (which equals e1) and e-1 as infinite sums, using the equation you have given.

    Cheers -- sylas
  4. Dec 19, 2009 #3


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    Homework Helper

    have you treid working backwards from the answer to understand how they get there?
  5. Dec 20, 2009 #4


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    Perhaps more to the point
    [tex]cos(x)= \sum_{n=0}^\infty \frac{(-1)^nx^n}{(2n)!}[/tex]

    What x gives [itex](-1)^nx^n= (-x)^n= 1[/itex]?

  6. Dec 20, 2009 #5


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    Um, actually
    [tex]cos(x)= \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}[/tex]
    This is not going to help. And let's not introduce the hyperbolic cos. He's got the answer, and now just needs to work back from there.

    Cheers -- sylas
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