Question

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

Answer 1

The series:

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

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

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)!)) )`

`abs (a_(n+1)/a_n) = abs( x^(n+1)/x^n) ( (2(n+1))!) / ((2n)!) (n^2!)/((n+1)^2!)`

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