Check whether the following series is uniformly convergent OR not?

Question

Check whether the following series is uniformly convergent OR not?

$\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}, x\in [0,\infty).$

My attempt

$\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}=x^4\sum_{n=1}^\infty \frac{1}{(1+x^2+x^4)^n}=( \frac{1}{(1+x^2+x^4)}).(\frac{x^4}{1-(\frac{1}{1+x^2+x^4})})=\frac{x^4}{(x^2+x^4)}$

$|\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}-\frac{x^4}{(x^2+x^4)}.1|=|\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}-\frac{x^4}{(x^2+x^4)}.\sum_{n=1}^\infty \frac{1}{2^n}|=|\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}-\sum_{n=1}^\infty \frac{x^4}{(x^2+x^4)}. \frac{1}{2^n}|=|\sum_{n=1}^\infty (\frac{x^4}{(1+x^2+x^4)^n}-\frac{x^4}{(x^2+x^4)}. \frac{1}{2^n})|$ How do I proceed further? Is there any theorem helps to solve it faster? I couldn't apply weirstrauss M-Test.

Answer 1

Hint

$$\sum_{n=1}^\infty \frac{x^4}{(1+x^2+x^4)^n}=\frac {x^2}{1+x^2}$$

Answer 2

By generalizing your first line, one can show the tail sum equals $$\sum_{n=N+1}^\infty \frac{x^4}{(1+x^2+x^4)^n} = \frac{x^4}{(1+x^2+x^4)^{N+1}}\frac{1}{1-\frac{1}{1+x^2+x^4}} = \frac{x^2}{1+x^2} \cdot \frac{1}{(1+x^2+x^4)^N}.$$

To show uniform convergence, it suffices to check that the following holds. $$\sup_{x \ge 0} \left(\frac{x^2}{1+x^2} \cdot \frac{1}{(1+x^2+x^4)^N}\right) \overset{N \to \infty}{\longrightarrow} 0.$$