`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 .