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Finding an expression for the series

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data
    1+2x+(4!x^(2))/((2!)^(2))+(6!x^(3))/((3!)^(2))+(8!x^(4))/((4!)^(2))+.....


    2. Relevant equations

    Uhhhhh...

    3. The attempt at a solution

    Well, the denominator looks like it's going to be (n!)^2. And this expression appears to be using values of N≥0. But I'm not so sure as to what the numerator may look like.
     
  2. jcsd
  3. Dec 7, 2011 #2

    Char. Limit

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    Just gonna import my answer over to here...


     
  4. Dec 7, 2011 #3
    Yea. Sorry about the screw-up of the thread I made. Can't type today, I guess. But thanks. Makes since.
     
  5. Dec 7, 2011 #4
    Alright. I think I got it. The true expression should be (2n!x^(n))/((n!)^(2))
     
  6. Dec 7, 2011 #5

    Char. Limit

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    That'll do it! Now just ratio-test that and simplify, simplify, simplify. You should get your answer before too long.
     
  7. Dec 7, 2011 #6
    Uh, yea. Still not getting a radius of convergence of (1/4). Could be an error in the book, or I'm doing something wrong. I think I'll ask my professor about this tomorrow.
     
  8. Dec 7, 2011 #7

    Char. Limit

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    What DO you get? If you show your work, I'm sure I can help you.
     
  9. Dec 7, 2011 #8
    Okay. I when I do the ratio test, it comes out to be the following...

    (2(n+1)!x^(n+1))/((n+1)!^(2)) * (n!^(2))/(2n!x^(n))

    this appears to equal....

    (xn!)/((n+1)!)

    But the limit of this as n→∞ equals zero. Still don't see how I'll get an answer of (1/4).
     
  10. Dec 7, 2011 #9

    Char. Limit

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    I think the problem is that you're seeing 2n! and 2(n+1)! as 2(n!) and 2((n+1)!), when you should be using (2n)! and (2(n+1))!. At least, that's the mistake the computer program I'm using to check my work is having. Fix that and you should be fine.
     
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