# More sequence problems :(

• May 16th 2008, 12:02 PM
simplysparklers
More sequence problems :(
Hiya guys!
I'd really appreciate some help with this question, I don't even know where to begin!This book I'm using is rubbish!

Consider the sequence $(a_{n})$, where
$a_{1}=0$, $a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}$ $(n\geq1)$

a.) Show by induction that $0\leq{a_{n}}<1$

b.) Show that $a_{n}$ is monotonic increasing

But where do I go about getting $a_{n}$ in the first place?? I'd really appreciate a full explaination. Thanks so much guys!

Jo
• May 16th 2008, 12:04 PM
Moo
Quote:

Originally Posted by simplysparklers
Hiya guys!
I'd really appreciate some help with this question, I don't even know where to begin!This book I'm using is rubbish!

Consider the sequence $(a_{n})$, where
$a_{1}=0$, $a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}$ $(n\geq1)$

a.) Show by induction that $0 \leq a_{n} <1$

b.) Show that $a_{n}$ is monotonic increasing

But where do I go about getting $a_{n}$ in the first place?? I'd really appreciate a full explaination. Thanks so much guys!

Jo

Fixed it :)

< or > is enough, you don't need any \ (Wink)
• May 16th 2008, 12:09 PM
simplysparklers
Lol!Cool!Thanks Moo!!I'm still only getting the hang of it!! :D
• May 16th 2008, 12:12 PM
Moo
For question a :

Basis : $0 \leq a_1=0 < 1$

Induction hypothesis : $0 \leq a_n <1$

(--> is $0 \leq a_{n+1} <1$ ?)

According to the hypothesis, $3 \leq a_n+3<4$

--> $\frac 13 \geq {\color{blue}\frac{1}{a_n+3}} > \frac 14 \Longleftrightarrow \frac 14 < {\color{blue}\frac{1}{a_n+3}} \leq \frac 13$ (for more convenience :))

Also according to the hypothesis, $0 \leq 3a_n <3$
$1 \leq {\color{red}3a_n+1} < 4$

---> $?? < \underbrace{({\color{red}3a_n+1}) \cdot {\color{blue}\frac{1}{a_n+3}}}_{a_{n+1}} < ??$

Can you continue ?

Quote:

Originally Posted by simplysparklers
Lol!Cool!Thanks Moo!!I'm still only getting the hang of it!! :D

No problem :D
You'll get used to it by practicing ^^

Edit : Ok, I didn't check it at first time, but my last inequality doesn't yield anything interesting :(
• May 17th 2008, 07:28 AM
simplysparklers
Quote:

Originally Posted by Moo
---> $?? < \underbrace{({\color{red}3a_n+1}) \cdot {\color{blue}\frac{1}{a_n+3}}}_{a_{n+1}} < ??$

Can you continue?

Edit : Ok, I didn't check it at first time, but my last inequality doesn't yield anything interesting :(

Hey Moo!
Are you saying then that this doesn't actually prove that $0\leq{a_n}<1$??
Because this: $?? < \underbrace{({\color{red}3a_n+1}) \cdot {\color{blue}\frac{1}{a_n+3}}}_{a_{n+1}} < ??$
is actually:
$0\leq{a_{N+1}}<1$, isn't it??

And any ideas on part b then anyone??Thanks for your help!! :)
• May 17th 2008, 07:30 AM
Moo
Quote:

Originally Posted by simplysparklers
Hey Moo!
Are you saying then that this doesn't actually prove that $0\leq{a_n}<1$??
Because this: $?? < \underbrace{({\color{red}3a_n+1}) \cdot {\color{blue}\frac{1}{a_n+3}}}_{a_{n+1}} < ??$
is actually:
$0\leq{a_{N+1}}<1$, isn't it??

And any ideas on part b then anyone??Thanks for your help!! :)

Actually, I wanted you to fill in the "??" gaps, with the previous inequalities, but it doesn't yield 0 and 1 ~

For b), show that $a_{n+1}>a_n$, this has to do with a quadratic inequation :)
• May 17th 2008, 07:55 AM
simplysparklers
So is the answer to the first one: $\frac{1}{4}\leq{(3a_{n}+1).\frac{1}{a_{n}+3}}<\fra c{4}{3}$
..which holds true, therefore, by induction, $0\leq{a_{n}}<1$ fpr all n?

And could you elaborate on part b please?I get that $(a_{n})$ is monotonic increasing if $a_{n+1}\geq{a_{n}}$ for all $n\geq{1}$, but how do you prove this?When you don't know what $(a_{n})$ is??
Coz like that, the last part of this questionis: Find $\lim_{n\rightarrow\infty}a_{n}$. I think I could do this part if I just knew where to start with getting $a_{n}$
I know I'm probably being really stupid about this, sorry!
• May 17th 2008, 08:49 AM
Isomorphism
Quote:

Originally Posted by simplysparklers
Hiya guys!
I'd really appreciate some help with this question, I don't even know where to begin!This book I'm using is rubbish!

Consider the sequence $(a_{n})$, where
$a_{1}=0$, $a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}$ $(n\geq1)$

a.) Show by induction that $0\leq{a_{n}}<1$

b.) Show that $a_{n}$ is monotonic increasing

But where do I go about getting $a_{n}$ in the first place?? I'd really appreciate a full explaination. Thanks so much guys!

Jo

Proving $a_n \geq 0$ by induction is almost trivial.

For $a_n < 1$,by induction, do the following:

$a_{n+1} = 3 - \frac{8}{a_{n}+3}$

$a_n < 1 \Leftrightarrow a_{n}+3 < 4 \Leftrightarrow -\frac1{a_{n}+3} < -\frac14 \Leftrightarrow a_{n+1} = 3-\frac8{a_{n}+3} < 1$
• May 17th 2008, 01:18 PM
simplysparklers
Thank you Isomorphism! :D I get the final line, but to begin with, where did you get $a_{n+1}=3-\frac{8}{a_{n}+3}$ in the first place??Is it really obvious and I'm just not seeing it?? (Thinking) My brain seems to work extra slow on sequences (Tongueout)
• May 17th 2008, 01:34 PM
Mathstud28
Quote:

Originally Posted by simplysparklers
Thank you Isomorphism! :D I get the final line, but to begin with, where did you get $a_{n+1}=3-\frac{8}{a_{n}+3}$ in the first place??Is it really obvious and I'm just not seeing it?? (Thinking) My brain seems to work extra slow on sequences (Tongueout)

$\frac{3a_n+1}{a_n+3}=\frac{3a_n+9-8}{a_n+3}=\frac{3a_n+9}{3a_n+3}-\frac{8}{a_n+3}=\frac{3(a_n+3)}{a_n+3}-\frac{8}{a_n+3}=3-\frac{8}{a_n+3}$