Numerical Sequence exercise for SV and SG - Leb_Edu_System

Exercise:

Consider a numerical sequence ${{\left( {{u}_{n}} \right)}_{n>0}}$ defined to be ${{u}_{0}}=5\,\,\,\And \,\,{{u}_{n+1}}=\frac{4{{u}_{n}}-2}{{{u}_{n}}+1}$

1) Show that, ${{u}_{n+1}}=4-\frac{6}{{{u}_{n}}+1}$ , $\forall n\in \mathbb{N}$

2) Deduce that, $2<{{u}_{n}}$  ,  $\forall n\in \mathbb{N}$ using Induction

3) Show that, ${{u}_{n+1}}-{{u}_{n}}=\frac{\left( 1-{{u}_{n}} \right)\left( {{u}_{n}}-2 \right)}{{{u}_{n}}+1}\,\,\,\,\,,\,\,\forall n\in \mathbb{N}$

4) Deduce that, ${{\left( {{u}_{n}} \right)}_{n>0}}$ is decreasing sequence

5) Let ${{v}_{n}}=\frac{{{u}_{n}}-2}{{{u}_{n}}-1}\,\,\,\,\,,\,\,\,\,\forall n\in \mathbb{N}$

i) Show that, ${{\left( {{v}_{n}} \right)}_{n}}$ is Geometric sequence with ratio $2/3$

ii) Write $\left( {{v}_{n}} \right)$ in terms of $n$ and $\left( {{u}_{n}} \right)$ in terms of ${{v}_{n}}$

iii) Deduce that, $\left( {{u}_{n}} \right)$ in terms of ${{v}_{n}}$

iv) Find the value of $\underset{n\to \infty }{\mathop{\lim }}\,{{u}_{n}}$

Solution:

1) ${{u}_{n+1}}=\frac{4{{u}_{n}}-2}{{{u}_{n}}+1}=\frac{4{{u}_{n}}-6+4}{{{u}_{n}}+1}=\frac{4\left( {{u}_{n}}+1 \right)-6}{{{u}_{n}}+1}=4-\frac{6}{{{u}_{n}}+1}$

2) For $n=0$ then ${{u}_{0}}=5>2$ true

Now suppose its true for ${{u}_{n}}>2$

We have ${{u}_{n}}>2\Leftrightarrow 2{{u}_{n}}>4\Leftrightarrow 4{{u}_{n}}-2{{u}_{n}}>2+2\Leftrightarrow 4{{u}_{n}}-2{{u}_{n}}-2>2$

So $4{{u}_{n}}-2>2{{u}_{n}}+2\Leftrightarrow 4{{u}_{n}}-2>2\left( {{u}_{n}}+1 \right)\Leftrightarrow \frac{4{{u}_{n}}-2}{{{u}_{n}}+1}>2\Leftrightarrow {{u}_{n+1}}>2$

3) ${{u}_{n+1}}-{{u}_{n}}=4-\frac{6}{{{u}_{n}}+1}-{{u}_{n}}=\frac{4{{u}_{n}}+4-6-{{u}_{n}}\left( {{u}_{n}}+1 \right)}{{{u}_{n}}+1}=\frac{4{{u}_{n}}-2-{{u}^{2}}_{n}-{{u}_{n}}}{{{u}_{n}}+1}$

             $=\frac{-{{u}^{2}}_{n}+3{{u}_{n}}-2}{{{u}_{n}}+1}=\frac{\left( 1-{{u}_{n}} \right)\left( {{u}_{n}}-2 \right)}{{{u}_{n}}+1}$

4) Using the result in part (2) we have $2<{{u}_{n}}\Leftrightarrow 0<{{u}_{n}}-2\,\,\,,\,\,\,3<{{u}_{n}}+1\,\,\,\,\And \,\,1-{{u}_{n}}<0$

So ${{u}_{n+1}}-{{u}_{n}}=\frac{\left( <0 \right)\left( >0 \right)}{>0}=\frac{<0}{>0}\le 0$

Therefore, $\left( {{u}_{n}} \right)$ is decreasing sequence

5)    ${{v}_{n}}=\frac{{{u}_{n}}-2}{{{u}_{n}}-1}$

i) ${{v}_{n+1}}=\frac{{{u}_{n+1}}-2}{{{u}_{n+1}}-1}=\frac{\frac{4{{u}_{n}}-2}{{{u}_{n}}+1}-2}{\frac{4{{u}_{n}}-2}{{{u}_{n}}+1}-1}=\frac{4{{u}_{n}}-2-2{{u}_{n}}-2}{4{{u}_{n}}-2-{{u}_{n}}-1}=\frac{2{{u}_{n}}-4}{3\left( {{u}_{n}}-1 \right)}$

So $r=\frac{{{v}_{n+1}}}{{{v}_{n}}}=\frac{\frac{2\left( {{u}_{n}}-2 \right)}{3\left( {{u}_{n}}-1 \right)}}{\frac{{{u}_{n}}-2}{{{u}_{n}}-1}}=\frac{2\left( {{u}_{n}}-2 \right)\left( {{u}_{n}}-1 \right)}{3\left( {{u}_{n}}-1 \right)\left( {{u}_{n}}-2 \right)}=\frac{2}{3}$

Therefore, $\left( {{v}_{n}} \right)$ is geometric sequence

ii) ${{v}_{0}}=\frac{{{u}_{0}}-2}{{{u}_{0}}-1}=\frac{5-2}{5-1}=\frac{3}{4}$

Since $\left( {{v}_{n}} \right)$ is geometric sequence with ${{v}_{0}}=\frac{3}{4}\,\,\,\And \,\,r=\frac{2}{3}$

Then ${{v}_{n}}={{v}_{0}}{{r}^{\left( n-0 \right)}}=\left( \frac{3}{4} \right){{\left( \frac{2}{3} \right)}^{n}}=\frac{3\times {{2}^{n}}}{{{2}^{2}}\times {{3}^{n}}}=\frac{{{2}^{n-2}}}{{{3}^{n-1}}}$

${{v}_{n}}=\frac{{{2}^{n-2}}}{{{3}^{n-1}}}\,\,\,\,\,,n\ge 0$

${{v}_{n}}=\frac{{{u}_{n}}-2}{{{u}_{n}}-1}\Leftrightarrow {{v}_{n}}{{u}_{n}}-{{v}_{n}}={{u}_{n}}-2\Leftrightarrow {{v}_{n}}{{u}_{n}}-{{u}_{n}}={{v}_{n}}-2\Leftrightarrow {{u}_{n}}=\frac{{{v}_{n}}-2}{{{v}_{n}}-1}$

iii) So ${{u}_{n}}=\frac{\frac{{{2}^{n-2}}}{{{3}^{n-1}}}-2}{\frac{{{2}^{n-2}}}{{{3}^{n-1}}}-1}=\frac{\frac{{{2}^{n-2}}-2\left( {{3}^{n-1}} \right)}{{{3}^{n-1}}}}{\frac{{{2}^{n-2}}-{{3}^{n-1}}}{{{3}^{n-1}}}}=\frac{{{2}^{n-2}}-2\left( {{3}^{n-1}} \right)}{{{2}^{n-2}}-{{3}^{n-1}}}$

iv) $\underset{n\to \infty }{\mathop{\lim }}\,{{u}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{2}^{n-2}}-2\left( {{3}^{n-1}} \right)}{{{2}^{n-2}}-{{3}^{n-1}}}=\frac{{{3}^{n-1}}\left( \frac{{{2}^{n-2}}}{{{3}^{n-1}}}-2 \right)}{{{3}^{n-1}}\left( \frac{{{2}^{n-2}}}{{{3}^{n-1}}}-1 \right)}$

Let $w\left( n \right)=\frac{{{2}^{n-2}}}{{{3}^{n-1}}}$ as $n\to \infty \,\,,\,\,w\to 0$

Remark:

$\underset{n\to \infty }{\mathop{\lim }}\,w\left( n \right)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{2}^{n-2}}}{{{3}^{n-1}}}=\infty \times 0\,\,\,ind\,\,form$

We have $w\left( n \right)={{2}^{n-2}}\times {{3}^{1-n}}$

$\ln w\left( n \right)=\ln \left( {{2}^{n-2}} \right)+\ln \left( {{3}^{1-n}} \right)=\left( n-2 \right)\ln 2+\left( 1-n \right)\ln 3$

$=n\ln 2-2\ln 2+\ln 3-n\ln 3=n\left( \ln \frac{2}{3} \right)-\left( 2\ln \frac{2}{3} \right)=\ln \frac{2}{3}\left( n-2 \right)$

But $\frac{2}{3}<0\Leftrightarrow \ln \frac{2}{3}\left( n-2 \right)<0$ so $w\xrightarrow[n\to \infty ]{}-\infty $

Thus $\underset{n\to \infty }{\mathop{\lim }}\,w\left( n \right)={{e}^{\underset{n\to \infty }{\mathop{\lim }}\,\ln w\left( n \right)}}={{e}^{-\infty }}=0$

Therefore, $\underset{n\to \infty }{\mathop{\lim }}\,{{u}_{n}}=\underset{w\to 0}{\mathop{\lim }}\,\frac{w-2}{w-1}=2$