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
If you know the common ratio of the geometric series ($r$) satisfies $-1<r<1$ you can use that fact to say "therefore it converges", the proof is short so I'll do it here.

Consider:

$S = 1+x+x^2+x^3+\dots+x^k+\dots+x^n$

then:

$xS= x+x^2+x^3+\dots+x^{k+1}+\dots+x^{n+1}$

Now let us subtract these, either way works:

$S-xS = 1-x^{n+1}$

So:

$S(1-x)=1-x^{n+1}$

Thus:

$S=\frac{1-x^{n+1}}{1-x}$

Which we all know and love, but look at that $x^{n+1}$, we have three cases, if $|x|<1$ then whatever $x$ is $x^k$ will have a smaller magnitude (will alternate in sign for negative x).

If $|x|>1$ then $|x\*x|>|x|$ so $x^k$ gets smaller in magnitude as k gets bigger.

IF $|x|=1$ we must inspect further, it could go either way.

So if $|x|<1$ then $x^{n+1}\rightarrow 0$ as $n\rightarrow$ infinity.

thus we have shown if $|x|<1$ then our geometric series for which x is the common ratio converges (because $x^{n+1}$ is the only part that changes with n)