Radius of convergence

I need to find the radius of convergence of the following function
and explanation of how you got the answer would be nice also.

Sigma of n=0 to infinity of [x^(2n+1)]/[(2n+1)!]

using ratio test :

∞
Σ [ x^(2n+1) / (2n+1)! ]
n = 0

(an + 1) / (an)

[ x^(2(n+1)+1) / (2(n+1)+1)! ] ÷ [ x^(2n+1) / (2n+1)! ]

[ x^(2n+2+1) / (2n+2+1)! ] * [ (2n+1)! / x^(2n+1) ]

[ x^(2n+3) / (2n+3)! ] * [ (2n+1)! / x^(2n+1) ]

[ x^(2n) * x^3) / (2n+3) * (2n+2) * (2n+1)! ] * [ (2n+1)! / x^(2n) * x^1) ]

[ x^2 / (2n+3) * (2n+2) ]

lim x^2 * [ 1 / (2n+3) * (2n+2) ] =
n-->∞

lim x^2 * [ 1 / (4n^2 + 10n + 6) ] = ====>large in charge
n-->∞

lim x^2 * [ 1 / (4n^2) ] = x^2 * [ 1 / (4*∞^2) ] = x^2 * (1 / ∞) = x^2 * 0 = 0 < 1
n-->∞

the radius of convergence is ∞

Use the ratio test.

r = lim(n→∞) |[x^(2n+3)/(2n+3)!] / [x^(2n+1)/(2n+1)!]
..= |x|^2 * lim(n→∞) 1/[(2n+2)(2n+3)]
..= 0.

Since r = 0 < 1 for all x, the series converges for all x.
(Hence the radius of convergence is infinite.)

I hope this helps!