<s>[INST] Tried to apply the ratio test to determine the convergence interval, but get the limit as a constant. So here is the series: $\Sigma_{n=1}^{\infty} \frac{x^{2n}}{1+x^{4n}}$
$$\left| \frac{x^{2(n+1)} (1+x^{4n})}{x^{2n}(1+x^{4(n+1)})}\right| = \left|\frac{x(1+x^{4n})}{1+x^{4n+4}} \right| \overset{\text{ n } \rightarrow \infty }{\rightarrow}\left[\frac{\infty}{\infty} \right]= \frac{x+x^{4n+1}}{1+x^{4n+4}} = \lim_{n \to \infty} \frac{x}{1+x^{4n+4}} +\lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}} = \\ 0 + \lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}} = \frac{\infty}{\infty} =\left[\text{ apply L'Hospital rule (4n+1)th times } \right] = \frac{1}{(4n+4)(4n+3)(4n+2)x^3} = 0 $$
Where is my mistake? Maybe some other technique would be useful for these series?
$$\lim_{n \to \infty} \frac{x}{1+x^{4n+4}} +\lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}}  \\ \text{ if } x = 0 \text{ then limit equals } 0 \\ \text{if x=1 then limit equals to 1 } \\ \text{if } 0<x<1 \text{ then limit is equal to x [ the same goes for } -1<x<0 \\ \text{ if } x> 1 \text{ then limit is equal to} \frac{1}{x^3} \text{ the same goes for x<1} $$
 [/INST] Without ratio test...
For $\vert x \vert \lt 1$ you have
$$0 \le \left\vert \frac{x^{2n}}{1+x^{4n}} \right\vert \lt \vert x \vert^{2n}$$ hence the series converges as $\sum x^{2n}$ converges.
For $\vert x \vert \gt 1$ you have
$$0 \le \left\vert \frac{x^{2n}}{1+x^{4n}} \right\vert \lt 1/\vert x \vert^{2n}$$ hence the series converges for the same reason.
If $x=1,-1$ the series diverges as its term is constant and equal to $1/2$.
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