Answer
The radius of convergence of the series is $0$.
Work Step by Step
Here, we have $a_n=\dfrac{(-1)^n (2n)! (x)^{2n}}{(n)!}$
Ratio Test states that when $\Sigma a_n$ is an infinite series with positive terms and, then $r=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_n}|$
a) When $0 \leq r \lt 1$, the series converges. (b) When $r \gt 1$, or, $\infty$, so the series diverges. (c) When $r=1$, the ratio test is inconclusive.
Now, $r=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(2n+2)! (x)^{2n+2}}{(n+1)!}}{\dfrac{(2n)! (x)^{2n}}{(n)!}}|\\=\lim\limits_{n \to \infty}|\dfrac{x^2(2n+1)(2n+2)}{n+1}|\\=\infty$
This implies that the radius of convergence of the series is $0$.
