Find the limit for this sequence

Question

Let $a \in \mathbb R$ and $x_0=a$ $$ x_{n+1}=3 -\frac{2}{x_n} $$

Find the limit for different values of $a$.

I see that if there is limit then it will be $1$ or $2$

Also, if $a=1$ then $(x_n)=(1,1,1,...) \to 1 $

if $a=2$ then $(x_n)=(2,2,2,...) \to 2 $

$ if a=0 \Rightarrow x_1 \notin \mathbb R \Rightarrow a \neq 0 $

if $ a > 2$

$x_2=3-\frac{2}{a}>3-1=2$

let $x_{n-1}>2$ then

$ x_n=3-\frac{2}{x_{n-1}} \ge 3-1=2$

then by induction $\forall n \in \mathbb N , x_n \ge2$

$\forall n \in \mathbb N $

$x_{n+1} -x_n= 3 -\frac{2}{x_n} -x_n= - \frac {{x_n}^2-3x_n+2}{x_n} =- \frac{(x-2)(x-1)}{x_n} < 0 $

Then $(x_n)$ is decreasing sequence. So the limit is exists and it's 2.

Also, if $1<a<2 $

$x_2=3-\frac{2}{a} \le 3-1=2$

let $x_{n-1} \le2$ then

$ x_n=3-\frac{2}{x_{n-1}} \le 3-1=2$

then by induction $\forall n \in \mathbb N , x_n<2$

$\forall n \in \mathbb N $

$x_{n+1} -x_n= 3 -\frac{2}{x_n} -x_n= - \frac {{x_n}^2-3x_n+2}{x_n}=- \frac{(x-2)(x-1)}{x_n} > 0 $

Then $(x_n)$ is increasing sequence. So the limit is exists and it's 2.

Now I need to find the limit for $a<1$

I tried to prove that but I couldn't. I tried to calculate some numerical examples for this case and I find the limit is 2 for theses examples and $ \exists n_0 \in \mathbb N : \forall n> n_0, (x_n)$ is a decreasing sequence and $ \forall n> n_0, x_n \ge 2 $

and its limit is 2. I tried to prove that but I can't. Any help?

Answer 1

Let $f(x)=3-\frac2x$, then $f$ has two distinct fixed points $y_1=1,y_2=2$.

If $a=2$, then $x_n=2,n=0,1,2,\cdots$. If $a=1$, then $x_n=1,n=0,1,2,\cdots$.

When $x_0=a\neq1,2$, Consider $$\frac{x_{n+1}-1}{x_{n+1}-2} =\frac{3 -\frac{2}{x_n}-1}{3 -\frac{2}{x_n}-2} =2\cdot\frac{x_n-1}{x_n-2},\quad n=0,1,2,\cdots.$$ we have $$\frac{x_n-1}{x_n-2}=2^n\cdot\frac{a-1}{a-2},\quad n=0,1,2,\cdots.$$ So $$x_n=\frac{2^{n+1}(a-1)-(a-2)}{2^n(a-1)-(a-2)},\quad n=0,1,2,\cdots.$$

Hence $\lim_{n\to\infty}x_n=2,\ \text{when}\ a\neq1$, and $\lim_{n\to\infty}x_n=1,$ when $a=1$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$ \begin{array}{c} \mbox{With}\quad a \in \mathbb{R}\quad \mbox{and}\quad x_{0} = a, \quad\color{#44f}{\large x_{n + 1} = 3 - 2/x_{n}}, \qquad {\large x_{n}}: {\large ?} \\ ------------------------------ \end{array} $$ \begin{align} & \mbox{Lets}\quad x_{n} \equiv {p_{n} \over q_{n}}\quad\mbox{with}\quad p_{0} = a,\ q_{0} = 1. \\[5mm] \mbox{I'll choose}\ {p_{n} \choose q_{n}} & = {3p_{n - 1} - 2q_{n - 1} \choose p_{n - 1}} = \pars{\begin{array}{rr} \ds{3} & \ds{-2} \\ \ds{1} & \ds{0} \end{array}}{p_{n - 1} \choose q_{n - 1}} \\[5mm] & = \pars{\begin{array}{rr} \ds{3} & \ds{-2} \\ \ds{1} & \ds{0} \end{array}}^{2}{p_{n - 2} \choose q_{n - 2}} = \cdots = \pars{\begin{array}{rr} \ds{3} & \ds{-2} \\ \ds{1} & \ds{0} \end{array}}^{n}{p_{0} \choose q_{0}} \\[5mm] & = \pars{\begin{array}{cc} \ds{2^{n + 1} - 1} & \ds{-2^{n + 1} + 2} \\ \ds{2^{n} - 1} & \ds{-2^{n} + 2} \end{array}}{a \choose 1} \\[5mm] & ={\bracks{2^{n + 1} - 1}a - 2^{n + 1} + 2 \choose \bracks{2^{n} - 1}a - 2^{n} + 2} \\[5mm] & \implies \color{#44f}{x_{n}} = \bbx{\color{#44f}{{\pars{2^{n + 1} - 1}a - 2^{n + 1} + 2\over \pars{2^{n} - 1}a - 2^{n} + 2}}} \\ & \end{align} How do I get the $\ds{n}$th-power of the matrix ?. There are, at least, two possible solutions:

$\ds{\left.\large 1\right)}$ Wolfram-${\tt Mathematica}$.

$\ds{\left.\large 2\right)}$ The matrix can be written as a linear combination of the Pauli Matrices and the $\ds{\quad\,\, 2 \times 2\ identity\ matrix}$. The $\ds{n}$th-power is found by extracting a coefficient of an $\,\,\,\quad$exponential generating function.

Answer 3

Answering this question, I wrote the steps for the solution of a first-order rational difference equation.

Using it, you would find that the solution is $$x_n=2+\frac{1}{2^n\,\frac{(a-1) }{(a-2)}-1}$$ which is the same as @Felix Marin result.