Show that the sequence is bounded below by ${\sqrt 2}$

Question

Find the limit of a sequence defined recursively as $x_1=\frac{3}{2}$, $x_{n+1}=\dfrac{4+3x_n}{3+2x_n}$ with $n\in \mathbb{N}$. Show that the limit exists before attempting to find it.

I need to show that it’s bounded below by ${\sqrt 2}$ , which I failed , but assuming that it is I was able to also show its decreasing so hence the existence of a limit. Any help on showing it’s bounded below by ${\sqrt 2}$ would be greatly appreciated . Thanks.

Answer 1

Expanding $x_{n+1}^2-2$ yields $$x_{n+1}^2-2=\frac{x_n^2-2}{(3+2x_n)^2}$$ Noting that $x_1 > \sqrt{2}$, it follows by induction on $n$ that $x_n > \sqrt{2}$ for all positive integers $n$.

Answer 2

Use induction. Assume $x_n > \sqrt{2}$. We need to show $x_{n + 1} > \sqrt{2}$.

$$x_n > \sqrt{2}$$ $$16 + 24x + 9x^2 > 18 + 24x + 8x^2$$ $$(4 + 3x)^2 > 2(3 + 2x)^2$$ $$\left(\frac{4 + 3x}{3 + 2x}\right)^2 > 2$$ $$(x_{n+1})^2 > 2$$ $$x_{n + 1} > \sqrt{2}$$

(formal proof goes top down, but it's much simpler to find it going down up)

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Set $\ds{x_{n} \equiv p_{n}/q_{n}}$ such that $\ds{p_{n + 1}/q_{n +1} = \pars{3p_{n} + 4q_{n}}/\pars{2p_{n} + 3q_{n}}}$.

Choose $\ds{p_{n + 1} = 3p_{n} + 4q_{n}}$ and $\ds{q_{n + 1} = 2p_{n} + 3q_{n}}$. Both equations can be rewritten as \begin{align} {p_{n + 1} \choose q_{n + 1}} & = \pars{\begin{array}{cc} \ds{3} & \ds{4} \\ \ds{2} & \ds{3} \end{array}} {p_{n} \choose q_{n}} = \pars{\begin{array}{cc} \ds{3} & \ds{4} \\ \ds{2} & \ds{3} \end{array}}^{2} {p_{n - 1} \choose q_{n - 1}} = \dots = \pars{\begin{array}{cc} \ds{3} & \ds{4} \\ \ds{2} & \ds{3} \end{array}}^{n} {p_{1} \choose q_{1}} \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, \pars{3 + 2\root{2}}^{n}{1 \over \root{3}}{\root{2} \choose 1} {1 \over \root{3}}\pars{\root{2}\quad 1}{p_{1} \choose q_{1}} \\[5mm] & = {\pars{3 + 2\root{2}}^{n} \over 3} \pars{\begin{array}{cc} \ds{2} & \ds{\root{2}} \\ \ds{\root{2}} & \ds{1} \end{array}} {p_{1} \choose q_{1}} \\[5mm] \implies & {p_{n + 1} \over q_{n + 1}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {2p_{1} + \root{2}q_{1} \over \root{2}p_{1} + q_{1}} = {2p_{1}/q_{1} + \root{2} \over \root{2}p_{1}/q_{1} + 1} = {3 + \root{2} \over 3\root{2}/2 + 1} \\[5mm] & = \bbx{\root{2}} \approx 1.4142 \end{align}