Determine the radius of convergence for the power series ∑n=1∞(n!)2(2n)!xn\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} x^n∑n=1∞(2n)!(n!)2xn.
R=1R = 1R=1
R=2R = 2R=2
R=4R = 4R=4
R=∞R = \inftyR=∞