# Convergence of a sequence using a auxliar sequence

Studying for a test, I had this question:

Given $$\alpha>0,$$ let a sequence $$(x_n)$$ given by: $$x_n=\frac{1}{\alpha+x_{n-1}}, x_1=\frac1{\alpha}$$ Proof that $$(x_{2k}), (x_{2k+1})$$ are monotone and find $$\lim x_n$$. Hint: do the analysis for $$(x_n-c),c=\frac{1}{a+c}>0$$ Following the hint, doing $$x_{2k+2}+c-x_{2k}+c=K(2c-x_{2k}), K>0$$

So, how can I prove $$2c

For me is clear that we want to show that the limit of the initial sequence is $$c.$$

Any help would be aprreciated.

Thanks for attention.

• I alteady noticed that $x_k>0$ and $2c>2-\alpha.$ Commented Apr 9, 2023 at 6:18

Here is another approach. \begin{align} x_{n+1}-x_{n-1} &=\frac1{a+x_n}-\frac1{a+x_{n-2}}\\ &=\frac{x_{n-2}-x_n}{(a+x_n)(a+x_{n-2})} \end{align} Thus, $$x_{n+1}-x_{n-1}$$ and $$x_n-x_{n-2}$$ have opposite signs. Likewise, $$x_{n+2}-x_n$$ and $$x_{n+1}-x_{n-1}$$ have opposite signs. Thus, $$x_{n+2}-x_n$$ and $$x_n-x_{n-2}$$ have the same sign. Since $$x_0=0$$ and $$x_2=\frac{a}{a^2+1}$$ and $$a\gt0$$, the even terms are increasing and the odd terms are decreasing.
Let $$c=\frac{-a+\sqrt{a^2+4}}2=\frac2{a+\sqrt{a^2+4}}$$, then $$c=\frac1{a+c}$$ and $$a+c=\frac{a+\sqrt{a^2+4}}2\gt1$$. Furthermore, \begin{align} x_n-c &=\frac1{a+x_{n-1}}-\frac1{a+c}\\ &=\frac{c-x_{n-1}}{(a+x_{n-1})(a+c)}\\ &=\frac{\frac1{a+c}-\frac1{a+x_{n-2}}}{\left(a+\frac1{a+x_{n-2}}\right)(a+c)}\\ &=\frac{x_{n-2}-c}{\underbrace{\left(a^2+ax_{n-2}+1\right)}_{\ge1+a^2}\underbrace{\,\left(a+c\vphantom{x_{n-2}}\right)^2\,}_{\ge(1+a/2)^2}} \end{align} Thus, $$x_n$$ is closer to $$c$$ than $$x_{n-2}$$, and on the same side of $$c$$, too. The convergence to $$c$$ is geometric.