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Show that two subsequences are monotonic and bounded

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hey again!! :)
Let the sequence $(a_{n})$ with $a_{1}>0$ and $a_{n+1}=1+\frac{2}{1+a_{n}}$.Show that the subsequences $a_{2k}$ and $a_{2k-1}$ are monotonic and bounded.Find the limit $\lim_{n \to \infty} a_{n}$,if it exists.
Do I have to show separately that the two subsequences are monotonic and bounded??Or is there an other way to show it??Could I for example show that $a_{n}$ is monotonic and bounded??
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: show that two subsequences are monotonic and bounded

Hey again!! :)
Let the sequence $(a_{n})$ with $a_{1}>0$ and $a_{n+1}=1+\frac{2}{1+a_{n}}$.Show that the subsequences $a_{2k}$ and $a_{2k-1}$ are monotonic and bounded.Find the limit $\lim_{n \to \infty} a_{n}$,if it exists.
Do I have to show separately that the two subsequences are monotonic and bounded?? Or is there an other way to show it?? Could I for example show that $a_{n}$ is monotonic and bounded??
They're probably hinting at a way to figure this out: find $a_{n+2}$ in terms of $a_{n}$. If you can reason from this expression adequately, you can kill both subsequences with one stone, to mix metaphors. The two subsequences they tell you to work with both have this in common: each term is two away from every other term in the original sequence.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Re: show that two subsequences are monotonic and bounded

They're probably hinting at a way to figure this out: find $a_{n+2}$ in terms of $a_{n}$. If you can reason from this expression adequately, you can kill both subsequences with one stone, to mix metaphors. The two subsequences they tell you to work with both have this in common: each term is two away from every other term in the original sequence.
I haven't understood.. (Worried) Could you explain it further to me??
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: show that two subsequences are monotonic and bounded

I haven't understood.. (Worried) Could you explain it further to me??
Ok, let's do this one thing at a time. Can you find $a_{n+2}$ in terms only of $a_{n}$?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Re: show that two subsequences are monotonic and bounded

Ok, let's do this one thing at a time. Can you find $a_{n+2}$ in terms only of $a_{n}$?
I found: $a_{n+2}=1+\frac{1}{1+\frac{1}{a_{n}}}$ .How can I continue?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: show that two subsequences are monotonic and bounded

I found: $a_{n+2}=1+\frac{1}{1+\frac{1}{a_{n}}}$ .How can I continue?
Hmm. That's not what I get:
$$a_{n+2}= \frac{3+2a_{n}}{2+a_{n}}.$$
Can you show your working?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Re: show that two subsequences are monotonic and bounded

Hmm. That's not what I get:
$$a_{n+2}= \frac{3+2a_{n}}{2+a_{n}}.$$
Can you show your working?
I tried it again and found the same result.. :)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Re: show that two subsequences are monotonic and bounded

I tried it again and found the same result.. :)
Do you mean the same result as you got before, or the same result that I got?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Re: show that two subsequences are monotonic and bounded

Do you mean the same result as you got before, or the same result that I got?
The same that you get!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Ah, so we're on the same page now. Can you compare $a_{n+2}$ to $a_{n}$ somehow? Maybe you can do $a_{n+2}-a_{n}$ or maybe $a_{n+2}/a_{n}$? If we can show this is monotonic and bounded, we'd be done with that part.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Ah, so we're on the same page now. Can you compare $a_{n+2}$ to $a_{n}$ somehow? Maybe you can do $a_{n+2}-a_{n}$ or maybe $a_{n+2}/a_{n}$? If we can show this is monotonic and bounded, we'd be done with that part.
I found $a_{n+2}-a_{n}=\frac{3-a_{n}^{2}}{2+a_{n}}$.