Find the interval of convergence of: sum(0 -> inf) [(3k)!/(2k)!] * x^k

Find the interval of convergence of:

sum(0 -> inf) [(3k)!/(2k)!] * x^k

I was trying to work out the ratio test and am getting something like

3/2 * |x| lim x->inf [(3k+1)*(3k+2)]/(2k+1).

Is that right? Do I need to do L'H rule, etc to eventually come to something for the radius of convergence? Thanks for your help!!

lim k->inf [(3(k+1))!/(2(k+1))!] * x^(k+1)] / [(3k)!/(2k)!] * x^k]
= lim k->inf |x|(3k+3)!(2k)! / [(2k+2)!(3k)!]
= lim k->inf |x|(3k)!(3k+1)(3k+2)(3k+3)(2k)! / [(2k)!(2k+1)(2k+2)(3k)!]
= lim k->inf |x|(3k+1)(3k+2)(3k+3) / [(2k+1)(2k+2)]
= lim k->inf |x|(3 + 1/k)(3 + 2/k)(3k+3) / [(2 + 1/k)(2 + 2/k)]
= [lim k->inf |x|(3k+3)] * {lim k->inf (3 + 1/k)(3 + 2/k) / [(2 + 1/k)(2 + 2/k)]}
= [lim k->inf |x|(3k+3)] * (3 + 0)(3 + 0) / [(2 + 0)(2 + 0)]
= [lim k->inf |x|(3k+3)] * (9/4)
= infinity if x is non-zero, but 0 if x = 0.

So the limit of the absolute value of the ratio of consecutive terms is greater than 1 for nonzero x, but less than 1 for x = 0.

It follows that the interval of convergence is just the single point x = 0. The radius of convergence is 0.

(Yes, you were on the right track, except it's slightly better to keep the |x| inside the limit this time, because of the somewhat unusual situation that occurs in the end ... the limit of the part not including the |x| ends up going to infinity. You could use L'Hopital's rule, but another way is to instead divide top and bottom by k. Dividing top and bottom by k or by an appropriate power of k is a common technique of finding the limit as k->inf of a rational function.)

Lord bless you today!