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[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.