Answer
The interval of convergence is $(-\infty,\infty)$.
Work Step by Step
We apply the ratio test
$$
\rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} |\frac{ x^{n+1}/((n+1)!)^2 }{ x^{n}/((n)!)^2 }|=|x|\lim _{n \rightarrow \infty} \frac{1}{(n+1)^2}=|x|\times0
$$
Hence, the series $\Sigma_{n=0}^{\infty} x^{n}/((n)!)^2$ converges if and only if $|x|\times0=0\lt1$, which is true for any value of $x$. That is, the interval of convergence is $(-\infty,\infty)$.
