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**hjortur** Let $\displaystyle x_0$ and $\displaystyle k$ be positive real numbers.

$\displaystyle

x_n = \frac{k}{1+x_{n-1}} , \text{ for } n\geq1

$

Now prove that either of the sequences $\displaystyle x_1 , x_3, x_5 \dotsb$ or $\displaystyle x_2 , x_4 , x_6 , \dotsb$ is increasing and the other one is decreasing.

**Then prove that **$\displaystyle x_1 , x_2 , \dotsb x_n$** is convergant.**

I have no idea how to work with recursive sequences, so if someone could tell me how do I start?

I will give a proof for the highlighted part, because I don't have any idea how to prove the first part yet, but I will write if I can find.

Using the recursion formula

$\displaystyle x(n)=\frac{k}{1+x(n-1)}$....for $\displaystyle n\geq1$,....**(1)**

where $\displaystyle x(0)>0$ and $\displaystyle k>0$, we learn that

$\displaystyle x(n)<k$....for all $\displaystyle n\geq1$, i.e., $\displaystyle \{x(n)\}_{n\in\mathbb{N}_{0}}$ is bounded.

Set

$\displaystyle \ell_{*}:=\liminf_{n\to\infty}x(n)$....and....$\displaystyle \ell^{*}:=\limsup_{n\to\infty}x(n)$.

Clearly,

$\displaystyle 0\leq\ell_{*}\leq\ell^{*}\leq k$.....**(2)**

Taking now inferior and superior limits on both sides of (1), we obtain

$\displaystyle \ell_{*}\geq\frac{k}{1+\ell^{*}}$....and....$\displaystyle \ell^{*}\leq\frac{k}{1+\ell_{*}}$ (actually they are equal, but I prefer inequalities for a general proof),

which yields

$\displaystyle \ell_{*}(1+\ell^{*})\geq k\geq\ell^{*}(1+\ell_{*})$,

or equivalently

$\displaystyle \frac{\ell_{*}}{1+\ell_{*}}\geq\frac{\ell^{*}}{1+\ ell^{*}}$.....**(3)**

Set $\displaystyle \nu(\lambda):=\lambda/(1+\lambda)$ for $\displaystyle \lambda>0$, and note that $\displaystyle \nu$ is increasing on $\displaystyle [0,\infty)$.

From (3), we have

$\displaystyle \nu(\ell_{*})\geq\nu(\ell^{*})$,

which yields

$\displaystyle \ell_{*}\geq\ell^{*}$.....**(4)**

It follows from (2) and (4) that $\displaystyle \ell_{*}=\ell^{*}$, i.e., $\displaystyle \{x(n)\}_{n\in\mathbb{N}_{0}}$ has a finite limit at infinity (more precisely this limit value lies in the interval $\displaystyle [0,k]$).

Let

$\displaystyle \ell:=\lim_{n\to\infty}x(n)$.

Taking limit on both sides of (1), we obtain

$\displaystyle \ell=\frac{k}{1+\ell}$,....**(5)**

and solving this we obtain

$\displaystyle \ell:=\frac{1}{2}\big(-1+\sqrt{1+4k}\big)$ (note that we took the positive solution of (5) since the solutions we consider are positive),

which is called the equilibrium of the recursive equation (1).

Every solution of (1) with positive initial value and positive $\displaystyle k$, tends to the equilibrium $\displaystyle \ell$.

You should refer to the following books for rational difference equations:

**[1]** M. R. S. Kulenovic and G. Ladas, *Dynamics of Second Order Rational Difference Equations with Open Problems and Conjectures*, Champman & Hall, 2002.

**[2]** E. Camouzis and G. Ladas, *Dynamics of Third-Order Rational Difference Equations with Open Problems and Conjectures*, Champman & Hall, 2008.

**Mathematica codes for plotting a graphic of the solution of the rational sequence**

Code:

s=30; "Number of iterations";
x=Table[0,{i,1,s+1}];
k=1; "Coefficient";
x[[1]]=1; "Initial Value";
For[i=1,i<=s,i++,x[[i+1]]=k/(1+x[[i]])]
eq=l/.Solve[l==k/(1+l),{l}];
eq=If[eq[[1]]>0,eq[[1]],eq[[2]]];
xtab=Table[{i,x[[i+1]]},{i,0,s}];
Show[ListPlot[xtab,PlotStyle->{Red,PointSize[0.02],Opacity[1]},PlotRange->Full,AxesOrigin->{0,0}],ListPlot[xtab,PlotStyle->{Blue,Dashed,Thickness[0.001],Opacity[1]},Joined->True],Plot[eq,{i,0,s},PlotStyle->{Green,Thickness[0.001]}]]