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 $2c<x_{2k}$

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.

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

1 Answer 1


Alternate Monotonicity

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.

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