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

sum_(n=0)^oo x^(2n)/((2n)!)
To find the 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) | (x^(2(n+1))/((2(n+1))!))/((x^(2n))/((2n)!))|
L=lim_(n->oo) | x^(2(n+1))/((2(n+1))!) * ((2n)!)/x^(2n)|
L= lim_(n->oo)| x^(2n+2)/((2n+2)!) * ((2n)!)/x^(2n)|
L= lim_(n->oo)| x^(2n+2)/((2n+2)*(2n+1)*(2n)!) * ((2n)!)/x^(2n)|
L=lim_(n->oo) |x^2/((2n+2)(2n+1))|
L=|x^2| lim_(n->oo)|1/((2n+2)(2n+1))|
L=|x^2| * 0
L =0
Take note that in Ratio Test, the series converges when L < 1.
Since the value of L is zero, which is less than 1, then the series converges for all values of x.
Therefore, the radius of convergence of the given series is R =oo .