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

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`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` .

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