Consider the power series f(x)=∑n=1∞(n!)2(2n)!xnf(x) = \sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} x^nf(x)=∑n=1∞(2n)!(n!)2xn. What is the radius of convergence RRR?
R=1R = 1R=1
R=2R = 2R=2
R=4R = 4R=4
R=∞R = \inftyR=∞