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

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Answer 1 · 2017-02-16T08:40:55

The series:

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

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

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#.