using the ratio test for the following
sum from n = 0 to infinity of $$ \sum_{m=0}^{+\infty}\frac{(-1)^m}{(m!)^2} x^{2m +10} $$
I need to find the radius of convergence. I managed to get up to $|x|^2$ $\lim$ as n approaches $∞$ of $\frac{-1(m!)}{m+1}$ but then what is the limit of that?
The series is dominated (in absolute values) by $$ x^{10}\sum_{m=0}^\infty \frac{(x^2)^m}{m!}=x^{10}e^{x^2} $$ so absolute convergence is ensured for all $x\in\Bbb R$.