Sequence exercise for Grade 11 Science ,Back 1

Exercise:

Let ${{\left( {{u}_{n}} \right)}_{n>0}}\,\,\And \,\,\,\,{{\left( {{v}_{n}} \right)}_{n>0}}$ be two sequences defined to be

         ${{u}_{1}}=\frac{3}{2}\,\,\,\,\And \,\,\,\,{{u}_{n+1}}=\frac{{{u}_{n}}+3}{2}$ and ${{v}_{n}}=3-{{u}_{n}}$ where $n\in \mathbb{N}$

a) Show that , ${{\left( {{v}_{n}} \right)}_{n>0}}$ is Geometric sequence with common ratio is half

b) Write ${{v}_{n}}$ in terms of $n$ then deduce ${{u}_{n}}$ in terms of $n$

c)  Show that, ${{\left( {{u}_{n}} \right)}_{n>0}}$ is increasing sequence

d) Let ${{S}_{n}}=\sum\nolimits_{i=1}^{n}{{{u}_{i}}\,\,\,}\And \,\,\,\,\,S_{n}^{*}=\sum\nolimits_{i=1}^{n}{{{v}_{i}}}$

      1)   Write ${{S}_{n}}$ in terms of $S_{n}^{*}$ , then deduce ${{S}_{n}}$ in terms of $n$

      2)    Find $\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}$

Solution:

a) $r=\frac{{{v}_{n+1}}}{{{v}_{n}}}=\frac{3-{{u}_{n+1}}}{3-{{u}_{n}}}=\frac{3-\frac{{{u}_{n}}+3}{2}}{3-{{u}_{n}}}=\frac{\frac{6-{{u}_{n}}-3}{2}}{3-{{u}_{n}}}=\frac{3-{{u}_{n}}}{2\left( 3-{{u}_{n}} \right)}=\frac{1}{2}$

Thus $\left( {{v}_{n}} \right)$ is geometric sequence with ratio ½

b) We already proved that $\left( {{v}_{n}} \right)$ is geometric sequence

hence the n-th term ${{v}_{n}}={{v}_{1}}{{r}^{n-1}}=\frac{3}{2}{{\left( \frac{1}{2} \right)}^{n-1}}=\frac{3}{{{2}^{n}}}$

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

thus ${{u}_{n}}={{v}_{n}}\left( {{2}^{n}}-1 \right)={{v}_{n}}{{M}_{n}}$ where ${{M}_{n}}={{2}^{n}}-1$ is Mersenne number

c) \[{{u}_{n+1}}-{{u}_{n}}=\frac{{{u}_{n}}+3}{2}-{{u}_{n}}=\frac{{{u}_{n}}+3-2{{u}_{n}}}{2}=\frac{-{{u}_{n}}+3}{2}=\frac{-\left( 3-{{v}_{n}} \right)+3}{2}=\frac{{{v}_{n}}}{2}=\frac{3}{{{2}^{n+1}}}>0\]

Hence ${{u}_{n+1}}>{{u}_{n}}\Leftrightarrow {{u}_{n}}<{{u}_{n+1}}$ thus ${{\left( {{u}_{n}} \right)}_{n>0}}$ is strictly increasing sequence

d) We have ${{v}_{n}}=3-{{u}_{n}}\Leftrightarrow \sum\nolimits_{i=1}^{n}{{{v}_{i}}=\sum\nolimits_{i=1}^{n}{\left( 3-{{u}_{i}} \right)=\sum\nolimits_{i=1}^{n}{3}-\sum\nolimits_{i=1}^{n}{{{u}_{i}}}}}$  why ??

1) Put $i=1\Leftrightarrow {{v}_{1}}=3-{{u}_{1}}$


Put $i=2\Leftrightarrow {{v}_{2}}=3-{{u}_{2}}$


Put $i=n\Leftrightarrow {{v}_{n}}=3-{{u}_{n}}$


Summing to get${{v}_{1}}+{{v}_{2}}+...+{{v}_{n}}=\underbrace{3+3+....+3}_{n-summand}-\left( {{u}_{1}}+{{u}_{2}}+....+{{u}_{n}} \right)$ \[\Rightarrow \sum\nolimits_{i=1}^{n}{{{v}_{n}}=3n-\sum\nolimits_{i=1}^{n}{{{u}_{n}}}\Leftrightarrow S_{n}^{*}=3n-{{S}_{n}}}\]

but $S_{n}^{*}$ represent the sum of geometric sequence hence

 $S_{n}^{*}=\frac{a\left( 1-{{r}^{n}} \right)}{1-r}=\frac{3}{2}\left( \frac{1-1/{{2}^{n}}}{1-1/2} \right)=3\left( 1-\frac{1}{{{2}^{n}}} \right)$

So \[{{S}_{n}}=3n-S_{n}^{*}=3n-3\left( 1-\frac{1}{{{2}^{n}}} \right)=3\left( n-1-\frac{1}{{{2}^{n}}} \right)\]

2) $\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=3\underset{n\to \infty }{\mathop{\lim }}\,\left( n-1-\frac{1}{{{2}^{n}}} \right)=\underset{n\to \infty }{\mathop{\lim }}\,n=\infty $ as $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{{{2}^{n}}}={{\left( \frac{1}{2} \right)}^{\infty }}=0$