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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

`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(2n+1)x^2|`

`L=|2x^2|lim_(n->oo) |2n+1|`

`L=|2x^2| * 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.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial