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

Feb 16, 2017

The series:

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

is absolutely convergent for $x \in \left(- \infty , + \infty\right)$

#### 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)!)

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = \left\mid x \right\mid \frac{\left(2 n + 2\right) \left(2 n + 1\right)}{\left({n}^{2} + 2 n + 1\right) \left({n}^{2} + 2 n\right) \cdot \ldots \cdot \left({n}^{2} + 1\right)}$

Clearly:

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = \left\mid x \right\mid \cdot {\lim}_{n \to \infty} \frac{\left(2 n + 2\right) \left(2 n + 1\right)}{\left({n}^{2} + 2 n + 1\right) \left({n}^{2} + 2 n\right) \cdot \ldots \cdot \left({n}^{2} + 1\right)} = 0$

for every $x$, thus the series is convergent for $x \in \mathbb{R}$.