Answer
The radius of convergence for the power series is $R=\infty$. This implies that the interval of convergence is $(-\infty, \infty)$.
Work Step by Step
Ratio Test; Let us consider an infinite series $\Sigma a_k$ with positive terms and suppose that $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|$
1) When $l \lt 1$. then series $\Sigma a_k$ converges absolutely.
2) When $l \gt 1$. then series $\Sigma a_k$ diverges.
3) When $l = 1$. then series $\Sigma a_k$ may converge or diverge, that is, the test is inconclusive.
Here, we have $a_k= \dfrac{k^2 x^{2k}}{k !}$
Now, $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty}|\dfrac{ \dfrac{(k+1)^2 x^{2(k+1)}}{(k+1)!}}{ \dfrac{k^2 x^{2k}}{k !}}|= |x| \lim\limits_{k \to \infty} \dfrac{k+1}{k^2}=0$
So, the radius of convergence for the power series is $R=\dfrac{1}{l}=\dfrac{1}{0}=\infty$. This implies that the interval of convergence is $(-\infty, \infty)$.
