Consider the MacLaurin series for #cos theta#:
#cos theta = sum_(n=0)^oo (-1)^n theta^(2n)/((2n)!)#
and let #theta = 3x^2#:
#cos (3x^2) = sum_(n=0)^oo (-1)^n(3x^2)^(2n)/((2n)!)#
#cos (3x^2) = sum_(n=0)^oo (-1)^n3^(2n)x^(4n)/((2n)!)#
Using the ratio test:
#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs ( ( 3^(2(n+1))x^(4(n+1))/((2(n+1))!))/(3^(2n)x^(4n)/((2n)!)))#
#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs (3^(2(n+1))/3^(2n) * ((2n)!) /((2(n+1))!)* x^(4(n+1))/x^(4n))#
#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo)3^(2n+2)/3^(2n) * ((2n)!)/((2n+2)!) * abs(x^(4n+4)/x^(4n))#
#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo)(9x^4)/((2n+1)(2n+2)) #
so for any value of #x in RR#:
#lim_(n->oo) abs (a_(n+1)/a_n) = 0#
and the series is convergent, which means that the radius of convergence is #R=oo#.
