Where Do These Two Power Series Converge on the Complex Plane?

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Homework Statement

sketch on the complex plane the region where the following two power series both converge

1) sigma from n=0 to infinity [(z-1)^n]/[n^2]

2) sigma from n=0 to infinity [((n!)^2)((z+4i)^n)]/[2n]!

The Attempt at a Solution

R=lim as n tends to infinity |(a(subscript n))/(a(subscript n+1))|

1) R=lim n tends to infinity [(z-1)^n][[n+1]^2]/[n^2][[z-1]^(n+1)]
=((n+1)^2)/(n^2)(z-1)
2) R=lim as n tends to infinity [((n!)^2)((z+4i)^n)/(2n)!][(2(n+1))!/(((n+1)!)^2)((z+4i)^(n+1))

I don't know how to proceed from here and I think I may have made a mistake somewhere

 

[HallsofIvy]

$$\frac{(n+1)^2}{n^2}= \frac{n^2+ 2n+ 1}{n^2}= 1+ \frac{2}{n}+ \frac{1}{n^2}$$
$\frac{(n!)^2}{((n+1)!)^2}\frac{((2n)!)^2}{((2n+1)!)^2}$$= \left(\frac{1}{n+1}\right)^2$$\left(\frac{1}{2n+1}\right)^2$