Here is the series: $\Sigma^{\infty}_{n=1}\frac{x^{2n}}{1+x^{4n}} $
Here is what I did: Given that $1+x^{4n} > x^{4n}$ we use the series $\sum^{\infty}_{n=1} \frac{x^{2n}}{x^{4n}} = \sum^{\infty}_{n=1} \frac{1}{x^{2n}}$, then for $|x| \ge 1 \ \ \sum^{\infty}_{n=1} \frac{1}{x^{2n}} \in R \Rightarrow \sum^{\infty}_{n=1} \frac{x^{2n}}{1+x^{4n}} \in R$
But in the feedback of this exercise, I was told that I haven't found the domain of convergence. From the previously given proof one can conclude that the domain of the convergence is $x \in (-\infty, -1] \cup [1, +\infty)$ or is it not correct? Furthermore, how can I test for uniform convergence within the domain of $[\ln3, + \infty)$? Ok, one poster gave a hint, that I need to determine the convergence domain for $\frac{1}{x^{2n}}$. So, in order for the series to convergence, I need that$-1< \frac{1}{x} < 1 $ (I think, I can take the geometric series power to be $2n$, correct?). Then for that to happen $|x|>1$. So, the domain of convergence for these geometric series is $(-\infty,-1[ \cup ]1,\infty)$. But I guess I cannot conclude that the domain of convergence for $\frac{x^{2n}}{1+x^{4n}}$ is the same, but how then I determine it?