What is the interval of convergence of sum_1^oo [(2n)!x^n] / ((n^2)! )1(2n)!xn(n2)!?

1 Answer
Feb 16, 2017

The series:

sum_(n=0)^oo ( (2n)!x^n)/((n^2)!)n=0(2n)!xn(n2)!

is absolutely convergent for x in (-oo,+oo)x(,+)

Explanation:

We can apply the ratio test, by evaluating:

abs (a_(n+1)/a_n) = abs( ( ( (2(n+1))!x^(n+1))/((n+1)^2!)) / ( ( (2n)!x^n)/((n^2)!)) )an+1an=∣ ∣ ∣(2(n+1))!xn+1(n+1)2!(2n)!xn(n2)!∣ ∣ ∣

abs (a_(n+1)/a_n) = abs( x^(n+1)/x^n) ( (2(n+1))!) / ((2n)!) (n^2!)/((n+1)^2!) an+1an=xn+1xn(2(n+1))!(2n)!n2!(n+1)2!

abs (a_(n+1)/a_n) = abs( x ) ((2n+2)!) / ((2n)!) (n^2!)/((n^2+2n+1)!) an+1an=|x|(2n+2)!(2n)!n2!(n2+2n+1)!

abs (a_(n+1)/a_n) = abs( x ) ((2n+2)(2n+1))/ ((n^2+2n+1)(n^2+2n)* ... * (n^2+1))

Clearly:

lim_(n->oo) abs (a_(n+1)/a_n) = abs(x)*lim_(n->oo) ((2n+2)(2n+1))/ ((n^2+2n+1)(n^2+2n)* ... * (n^2+1)) = 0

for every x, thus the series is convergent for x in RR.