Originally Posted by

**redsoxfan325** When I took the limit we already knew that it was monotonically decreasing and bounded below by 0.

I don't think so, since we don't know that $\displaystyle x(n)>\sqrt{\alpha}$ yet, we only know this for $\displaystyle n=1$, it should be repeated by induction.

Therefore the proof for the decreasing behaviour by **naught101 **is not actually a good one.

Here is what I have, I hope you will find it rigorous.

Let

$\displaystyle f(\lambda):=\frac{1}{2}\bigg(\lambda+\frac{\alpha} {\lambda}\bigg)$....for $\displaystyle \lambda\in(0,\infty)$.

So the rational difference equation can be written as

$\displaystyle x(n+1)=f(x(n))$....for $\displaystyle n\in\mathbb{N}$.....**(1)**

Then, we see that

$\displaystyle f^{\prime}(\sqrt{\alpha})=0$, $\displaystyle f(0)=f(\infty)=\infty$ and $\displaystyle f(\sqrt{\alpha})=\sqrt{\alpha}$,....**(2)**

i.e., $\displaystyle f$ attains its minimum value at $\displaystyle \sqrt{\alpha}$ with the minimum value $\displaystyle \sqrt{\alpha}$ which is at the same time the unique equilibrium of the rational difference equation, i.e., there is no value $\displaystyle \lambda\in(0,\infty)\backslash\{\sqrt{\alpha}\}$ such that $\displaystyle f(\lambda)=\lambda$.

Now, define $\displaystyle g(\lambda):=\lambda-f(\lambda)$....for $\displaystyle \lambda\in(0,\infty)$.

Here is the most important part.

Clearly, $\displaystyle g^{\prime}(\lambda)=(\lambda^{2}+\alpha)/(2\lambda^{2})>0$....for all $\displaystyle \lambda\in(0,\infty)$.

And $\displaystyle g(\sqrt{\alpha})=\sqrt{\alpha}-f\big(\sqrt{\alpha}\big)=0$, which yields $\displaystyle g(\lambda)>0$....for all $\displaystyle \lambda\in(\sqrt{\alpha},\infty)$, i.e.,

$\displaystyle \lambda>f(\lambda)>f(\sqrt{\alpha})=\sqrt{\alpha}$....for all $\displaystyle \lambda\in(\sqrt{\alpha},\infty)$.....**(3)**

See the following graphic.

Red: $\displaystyle \nu=\lambda$, Blue: $\displaystyle \nu=f(\lambda)$, and Green: $\displaystyle \nu=\sqrt{\alpha}$.

*Codes of the graphic for Mathematica 7.0*

Code:

Show[{Plot[{\[Lambda],1/2(\[Lambda]+1/\[Lambda]),1},{\[Lambda],0,3},PlotRange->{{0,3},{0,3}},PlotStyle->{{Red},{Blue},{Green}},AxesOrigin->{0,0},AxesLabel->{\[Lambda],\[Nu]},LabelStyle->Directive[White,Large]],Graphics[{Text[Style[Sqrt[\[Alpha]]],{2.7,0.8}],Text[Style["f"],{2.7,1.4}],Text[Style["I"],{2.7,2.5}]}]}]

Now, we are ready to finalize the proof.

Let $\displaystyle x(1)>\sqrt{\alpha}$, then taking $\displaystyle f$ on both sides by considering (3), we have

$\displaystyle x(1)>f\big(x(1)\big)=x(2)>\alpha$.....**(4)**

Then similarly considering (3) and (4), we have

$\displaystyle x(2)>f\big(x(2)\big)=x(3)>\alpha$, and by the emerging pattern, we have in general that

$\displaystyle x(1)>x(n)>x(n+1)>\alpha$....for all $\displaystyle n\in\mathbb{N}$,

which implies that the sequence is decreasing and so it has a finite limit, more precisely, $\displaystyle \ell\in[\sqrt{\alpha},x(1))$, where $\displaystyle \ell:=\lim\nolimits_{n\to\infty}x(n)$.

The rest is simple, passing now to limit as $\displaystyle n\to\infty$ on both sides of (1) and using (2), we learn that $\displaystyle \ell=\sqrt{\alpha}$.

This completes the solution.