Prove that the geometric series $1+\frac{4x}{5+x^2} + (\frac{4x}{5+x^2})^2+\cdot$ is convergent for all values of x..

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

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You ought to prove that, for all $x$, $$\left\lvert \frac{4x}{5+x^2}\right\rvert <1$$ Can you do that? – Pedro Tamaroff Aug 29 '13 at 22:46
What you are doing to show convergence is fine; if you multiply through by 4 and then bring terms to one side, you get the inequalities $x^2-4x+5>0$ and $x^2+4x+5>0$. Then you can complete the square to show these are valid for all x. – user84413 Aug 29 '13 at 22:52
@user84413 so if I just complete the square of find the minimum point and show that $x^2 -4x+5>0$, does that constitute to a proof? I haven't studied modulus functions at school yet so I don't know how to do it like the 3 answers below have. – salman Aug 29 '13 at 22:59
Right, you can use that $x^2-4x+5=(x-2)^2+1\ge1$ and $x^2+4x+5=(x+2)^2+1\ge1$ to show these are both positive. An alternative would be to show that the discriminant $b^2-4ac<0$, which implies that these expressions are always positive or always negative, and then just observe that they're positive at $x=0$, say. – user84413 Aug 29 '13 at 23:31
As to your question about simplification, one could express the answer as $1+\frac{4x}{5+x^2-4x}$. Hard to say whether this is "better." – André Nicolas Aug 30 '13 at 1:01

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

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I understood this part. With respect to finding the sum to infinity, is my answer correct? – salman Aug 29 '13 at 23:04
Yes it is, @user90771 – DonAntonio Aug 30 '13 at 7:22

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.

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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)

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