Given $a>0$, define inductively the sequence $(x_n)$ puting $x_1=1/a$ and $x_{n+1}=1/(a+x_n)$. Consider the number $c$, positive root of the equation $x^2+ax-1=0$, the only positive number that $c=\dfrac{1}{a+c}$. Prove that:
$$x_2<x_4<...<x_{2n}<...<c<...<...<x_{2n-1}<...<x_3<x_1$$
and that we have $\lim x_n=c$. The number $c$ could be consider the result of the sum of the continuous fraction:
$$\dfrac{1}{a+\dfrac{1}{a+\dfrac{1}{a+\dfrac{1}{a+...}}}}$$.
I have seen a proof from $x_2<x_4<...<x_{2n}<...<c<...<...<x_{2n-1}<...<x_3<x_1$, but it seens a little bit extensive demonstration, and I wonder if there's another way to proof, and my question is if it's possible to demonstrate using induction. I tried but I got stuck apparently because we have to involve the odd terms to proof $x_2<x_4<...$ and vice-versa.
