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 $x in RR$ .

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