sum_(n=0)^oo (2n)!x^(2n)/(n!) Find the radius of convergence of the power series.

sum_(n=0)^oo (2n)! x^(2n)/(n!)
To find radius of convergence of a series sum a_n , apply the Ratio Test. 
L = lim_(n->oo) |a_(n+1)/a_n|
L=lim_(n->oo)| ((2(n+1))! x^(2(n+1))/((n+1)!))/((2n)! x^(2n)/(n!))|
L=lim_(n->oo) | ((2n+2)!)/((2n)!) * (x^(2n+2)/((n+1)!))/(x^(2n)/(n!))|
L=lim_(n->oo) | ((2n+2)!)/((2n)!) * x^(2n+2)/((n+1)!)*(n!)/x^(2n)|
L= lim_(n->oo) | ((2n+2)(2n+1)(2n)!)/((2n)!) * x^(2n+2)/((n+1)n!)*(n!)/x^(2n)|
L=lim_(n->oo) | ((2n+2)(2n+1)x^2)/(n+1)|
L=lim_(n->oo)|(2(n+1)(2n+1)x^2)/(n+1)|
L=lim_(n->oo) |(2(2n+1)x^2|
L=|2x^2|lim_(n->oo) |2n+1|
L=|2x^2| * oo
L=oo
Take note that in Ratio Test,  the series diverges when L > 1.
So the series diverges except at x=0.
Since the series converges at x=0 only, therefore, the radius of convergence is R=0.