Answer
The radius of convergence of the series is $\infty$.
Work Step by Step
Here, we have $a_n=\dfrac{(-1)^n (x)^{2n}}{(2n)!}$
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, $r=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(x)^{2n+2}}{(2n+2)!}}{\dfrac{(x)^{2n}}{(2n)!}}|\\=\lim\limits_{n \to \infty}|\dfrac{x^2}{(2n+2)(2n+1)}|\\=0$
Therefore, the series converges for all the values of $x$. This implies that the radius of convergence of the series is $\infty$.
