Convergence of sequence $x_{n+1}=\frac{x_n^2+2}{2x_{n}-1}$

Question

I got a problem as follows, I have stucked for a week and I got no progress... May I recieve some guidance?

Define a sequence $\{x_n\}$ by $x_1=c$ and $\displaystyle x_{n+1}=\frac{x_{n}^{2}+2}{2x_n - 1}$ for $n\in\mathbb{N}$, where $c\neq \frac{1}{2}$ is a constant.

Prove that $\{x_n\}$ is convergent and find its limit.

(Hint: Consider $\displaystyle c>\frac{1}{2}$ and $\displaystyle c<\frac{1}{2}$ separately.)

I can kind of guess $\displaystyle\lim_{n\rightarrow\infty}x_n = 2$, and find the recurrence relationship between $|x_{n+1} - 2|$ and $|x_n - 2|$, but I failed to proceed. Nevertheless, I tried with the Cauchy convergence criterion by considering $|x_{n+1} - x_n|$ and I end up with nothing.

Thank you in advance for your kind reply.

Answer 1

Let $g$ denote a possible limit of the sequence $x_n.$ Then $$g={g^2+2\over 2g-1}$$ Hence $g=2$ or $g=-1.$

Consider the case $c>{1\over 2}.$ Assume $x_n>{1\over 2}.$ Then $x_{n+1}>2.$ Indeed $${x_n^2+2\over 2x_n-1}-2={(x_n-2)^2\over 2x_n-1}>0$$ Moreover $x_{n+1}<x_n,$ as $${x_n^2+2\over 2x_n-1}-x_n=-{(x_n-2)(x_n+1)\over 2x_n-1} \ (*)$$ Therefore the sequence $x_n$ is decreasing and bounded below by $2$ for $n\ge 2.$ Hence $x_n$ is convergent and $g=2.$

Let $c<{1\over 2}.$ By the similar procedure based on the formula $${x_n^2+2\over 2x_n-1}+1={(x_n+1)^2\over 2x_n-1}$$ and on $(*)$ we can prove that $x_n<-1,$ $x_{n+1}>x_n$ for $n\ge 2$ and the limit is equal $-1.$

Answer 2

This is NOT an answer, just some analysis

Take the subtraction:

$$x_{n+1}-x_n=\frac{-(x_n-2)(x_n+1)}{2x_n-1}$$

If we make an analogy with the autonomous differential equation, $$x'=f(x)=\frac{-(x-2)(x+1)}{2x-1}$$

we can see there are three critical points, where $x=-1$ and $x=2$ are stable, $x=\frac{1}2$ is a saddle point.

If intial value $x_0>\frac{1}2$ it will converges to $2$, and if initial value $x_0<\frac{1}2$, it will converge to $-1$.

Answer 3

Let $\,2 x_n - 1 = u_n \,$, then substituting $\,x_n = \dfrac{u_n+1}{2}\,$ in the original recurrence gives:

$$ \frac{u_{n+1}+1}{2} = \frac{u_n^2 + 2 u_n + 9}{4u_n} \quad\iff\quad u_{n+1} = \frac{u_n^2 + 9}{2u_n} \tag{1} $$

Substituting $\,u_n = 3v_n\,$ in $\,(1)\,$ gives: $$ 3v_{n+1} = \frac{9 v_n^2 + 9}{2 \cdot 3 v_n} = \frac{3}{2}\,\frac{v_n^2 + 1}{v_n} \quad\iff\quad v_{n+1} = \frac{1}{2}\left(v_n + \frac{1}{v_n}\right) \tag{2} $$

Then:

$$ \frac{v_{n+1}-1}{v_{n+1}+1} = \frac{\frac{1}{2}\left(v_n + \frac{1}{v_n}\right)-1}{\frac{1}{2}\left(v_n + \frac{1}{v_n}\right)+1} = \frac{v_n^2 - 2 v_n + 1}{v_n^2 + 2 v_n + 1} = \left(\frac{v_n-1}{v_n+1}\right)^2 = \ldots = \left(\frac{v_1 - 1}{v_1 + 1}\right)^{2^n} $$