Let $y = \prod_{n=0}^{\infty} \left ( 1 + \frac{1}{x^{2n}} \right ) \implies log(y) = \sum_{n=0}^{\infty} log \left ( 1 + \frac{1}{x^{2n}} \right )$

Now, let $f_n(x) = log \left ( 1 + \frac{1}{x^{2n}} \right ) $

Using ratio test,

For $0<|x|\leq1, \frac{f_{n_1}(x)}{f_n(x)} \geq 1$, and thus it's divergent (Equality holds for $x=1$)

For $1<|x|<\infty, \frac{f_{n_1}(x)}{f_n(x)} < 1$, and hence the series $ log(y) $ is convergent

Now, the value of $log(y)$ is greater than 0, so we can take anti-log,

$\implies$ the given series $\prod_{n=0}^{\infty} \left ( 1 + \frac{1}{x^{2n}} \right )$ is also convergent for $1<|x|<\infty$