Prove that sequence $ x_{(n+1)}= \frac {a}{1+x_n}$ is convergent to positive root of $x^2+x-a=0,$where $a >0$ and $x_1 >0$ Prove that sequence $$ x_{(n+1)}= \frac {a}{1+x_n}$$ is convergent to positive root of $x^2+x-a=0,$where $a >0$ and $x_1 >0$
we have $$x_{(n+2)}-x_n=\frac{-a(x_{(n+1)}-x_{(n-1)})}{(1+x_{(n-1)})(1+x_{(n+1)})}$$
Hence from this we see behaviour of even terms subsequence and odd terms subsequence  is opposite. how to proceed from there. any hint please
 A: Yes, the behavior is opposite indeed, but, as a continuation of your work, let's look at 
$$x_{n+2}=\frac{a}{1+x_{n+1}}=\frac{a}{1+\frac{a}{1+x_n}}=\frac{a(x_n+1)}{1+x_n+a}$$
or $x_{n+2}=f(x_n)$ where $f(x)=\frac{a(x+1)}{1+x+a}$. $f(x)$ is ascending because $f'(x)=\left(\frac{a}{1+x+a}\right)^2$. It is also worth mentioning that from $n=2$ we definitely have $0<x_n=\frac{a}{1+x_{n-1}}\leq a$ and $0< f(x)\leq a$ as well, for $x>0$. Thus, technically, we can treat $f(x)$ as a mapping $f:[0,a]\to[0,a]$. It is also a contraction map since, from MVT, $\forall x,y$ there $\exists\epsilon$ in between s.t.
$$|f(x)-f(y)|=|f'(\epsilon)|\cdot|x-y|<\left(\frac{a}{1+a}\right)^2\cdot|x-y|$$
with $q=\left(\frac{a}{1+a}\right)^2\in[0,1)$. As a result


*

*odd terms sub-sequence will converge ...

*even terms sub-sequence will converge ...


... to a $x>0$ that satisfies
$$x=\frac{a(x+1)}{1+x+a}\iff
ax+a=x+x^2+ax \iff\\
x^2+x-a=0$$
A: Let $\alpha=\frac{-1+\sqrt{1+4a}}{2}$ and then $\alpha$ is the positive root of $x^2+x-a=0$. Let $y_n=\frac{1}{x_n-\alpha}$ and then $x_n=\alpha+\frac{1}{y_n}$. The recursive relation becomes
$$ y_{n+1}=-\frac{1+\alpha}{\alpha}y_n-\frac{1}{\alpha}$$
which is linear and can be solved easily. In fact,
$$ y_n= c
   \left(-\frac{\alpha+1}{\alpha}\right)^{n-1}-\frac{1-\left(-\frac{\alpha+1}{\alpha}\right)^n}{2\alpha+1} $$
where $c$ is some constant. Now $y_n\to\infty$ as $n\to\infty$ and hence $x_n-\alpha\to0$ or $x_n\to\alpha$ as $n\to\infty$.
