What is the limit of the serie $\sum_{k=0}^{\infty}\frac{x^4}{(1+x^2)^k}$

Question

Find the limit of the series :

$$\sum_{k=0}^{\infty}\frac{x^4}{(1+x^2)^k}$$

So, I applied the ratio test and did $\frac{x^4}{(1+x^2)^{n+1}}$*$\frac{(1+x^2)^n}{x^4}$.

Then, I get $1 + \frac{1}{x^2}$. I then multiply $1 + \frac{1}{x^2}$ by $x^4$, to which I get an answer choice of (E) $x^4 + x^2$

I kind of want to know if I'm actually doing the process correctly, and that $x^4 + x^2$ is the limit of the aforementioned series.

These are the following answer choices:

(A) $x^6 + x^4$ (B) $\frac{x^6}{1+x^2}$ (C) $x^6$ (D) $x^4$ + $\frac{x^6}{1+x^2}$ (E) $x^4$ + $x^2$

Answer 1

hint: You basically have a geometric series with the first term $a_1 = x^4, r = \dfrac{1}{1+x^2}$. The sum is $\dfrac{a_1}{1-r}=...$

Answer 2

The sum is $x^4(\sum_{n\geq 0}({1\over{1+x^2}})^n=x^4({1\over{1-{1\over{1+x^2}}}})$