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Radius of convergence

  1. Mar 27, 2008 #1
    I am looking for radius of convergence of this power series:
    [tex]\sum^{\infty}_{n=1}[/tex]a[tex]_{n}[/tex]x[tex]^{n}[/tex], where a[tex]_{n}[/tex] is given below.
    a[tex]_{n}[/tex] = (n!)^2/(2n)!

    I am looking for the lim sup of |a_n| and i am having trouble simplifying it. I know the radius of convergence is suppose to be 4, so the lim sup should equal 1/4.

    here is my work (leaving lim sup as n-> inf out as i don't want to write it every line)
    [(n!)^2/(2n)!]^(1/n)
    I expanded out the bottom factorial:
    [(n!)(n!) / (2n)(2n-1)(2n-2)(2n-3)(2n-4)! ] ^ (1/n)
    and found that you can take out a 2 from the bottom even terms (now i don't know how to express the odd terms as a factorial):
    [(n!)(n)(n-1)(n-2)! / (2)(n)(n-1)(n-2)!(2n-1)(2n-3)(2n-5)...! ] ^ (1/n)
    and i canceled out the n! in top and bottom:
    [ (n!) / 2 (2n-1)(2n-3)(2n-5)(2n-7)..! ] ^ (1/n)
    now i am stuck..

    any help would be highly appreciated. haven't really dealt with factorials in awhile.
     
  2. jcsd
  3. Mar 27, 2008 #2

    HallsofIvy

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    The way you have written that makes it very difficult to read. I presume you are using the ratio test to find the radius of convergence.

    [tex]a_{n+1}|x^{n+1}|= \frac{((n+1)!)^2}{(2(n+1))!}|x^{n+1}|[/tex]
    and you want to divide that by
    [tex]a_n|x^n|= \frac{(n!)^2}{(2n)!}|x^n|[/tex]
    Okay, just be careful to combine the corresponding parts into fractions:
    [tex]\left(\frac{(n+1)!}{n!}\right)^2\frac{(2n)!}{(2n+2)!}|x|[/tex]
    Now, you certainly should know that (n+1)!/n! = n+1 and it is not to difficult to see that (2n+2)!= (2n+2)(2n+1)(2n)! so that (2n)!/(2n+2)!= 1/((2n+2)(2n+1)). What you have reduces to
    [tex]\frac{(n+1)^2}{(2n+2)(2n+1)}|x|[/tex]
    and since 2n+2= n+1 we can cancel n+1 in numerator and denominator to get
    [tex]\frac{n+1}{2(2n+1)}|x|[/tex]
    Now, what is the limit of that as n goes to infinity?
     
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