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

`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.

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