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
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
... to a $x>0$ that satisfies $$x=\frac{a(x+1)}{1+x+a}\iff ax+a=x+x^2+ax \iff\\ x^2+x-a=0$$
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 $n\to\infty$.
