Question: Prove that the geometric series $1+\frac{4x}{5+x^2} + (\frac{4x}{5+x^2})^2+ \cdot\cdot\cdot$ is convergent for all values of x and find the sum of an infinite number of terms of the series.

I've established that $a=1$ and $r=\frac{4x}{5+x^2}$ and that for a geometric to converge to a limit $-1 \lt r \lt 1$ as $n\to\infty$. So that $-1 \lt \frac{4x}{5+x^2} \lt 1$ $\therefore$ $\frac{-(5+x^2)}{4} \lt x \lt \frac{(5+x^2)}{4}$

Is this of any significance in proving that the series converges? I can't figure the next step.

With finding the sum to infinity, I got: $S\_\infty = \frac{1}{1-\frac{4x}{5+x^2}} = \frac{5+x^2}{5+x^2-4x}$. Can this be further simplified?

Thanks

Answer: 
A geometric series is convergent iff its constant quotient is less than one in absolute value, and in this cases iff

$$\left|\frac{4x}{5+x^2}\right|<1\iff|4x|<|5+x^2|\iff 16x^2<x^4+10x^2+25\iff$$

$$\iff x^4-6x^2+25>0\iff (x^2-5)^2+4x^2>0$$

And the last inequality is immediate since it is the sum of two non-negative numbers one of each (at least) is always positive
For every $x \ge 0$ we have
$$
x^2-4x+5=(x-2)^2+1>0,
$$
i.e.
$$
q(x):=\frac{4x}{x^2+5}<1.
$$
Using the fact that $q(-x)=-q(x)$ we get that
$$
-1<q(x)<1 \quad \forall x\in \mathbb{R}.
$$
Hence the series
$$
1+q(x)+q^2(x)+\ldots
$$
is convergent.