you can use lim n->oo ||a_n|| / ||a_(n+1)|| if the limit exists. otherwise you'll have to cauchy-hadamard or resort to other means.
for your first question for example, i would use cauchy hadamard.
the thing on the bottom of the denominator for cauchy hadamard is
limsup nth root of (log n)² = limsup (log n)^(2/n) = limsup [(log n)^(1/n)]^2
≤ lim [n^(1/n)]^2] = 1^2 = 1
so radius of convergence ρ = 1/1 = 1 where it's centred on the point (0,i)