Show that the sequence defined by $a_1=1 \text{ } a_{n+1}=3-\frac{1}{a_n}$ is increasing and $a_n

Question

Show that the sequence defined by $a_1=1 \text{   } a_{n+1}=3-\frac{1}{a_n}$ is increasing and $a_n<3$ for all n.  Deduce that $\left\{ a_n \right\}$ is convergent and find its limit.

anonymous · Answer

Sorry guys, I have no idea how to do this one...

Note, I could have typed it a little better:   it is defined as:  $$a_1=1 \text{  ,           } a_{n+1}=3-\frac{1}{a_n}$$

anonymous · Answer

assuming it has a limit, the limit is easy to find
set 
$$x=3-\frac{1}{x}$$ and solve for $x$

anonymous · Answer

it is more or less clear that if it is increasing, then $3-\frac{1}{a_n}<3$ as you are subtracting a positive number for 3
that means you have an increasing sequence, bounded above by 3, so it must have a limit
the hard part is showing that it is increasing

anonymous · Answer

*from 3

anonymous · Answer

Well the number that gets subtracted is getting smaller each time, so the overall value gets bigger each time

anonymous · Answer

ok we can do that one too

anonymous · Answer

you want to show that the sequence is  increasing, meaning that $a_{n+1}>a_n$ right?

anonymous · Answer

lets right down specifically what that means
since $a_{n+1}=3-\frac{1}{a_n}$ this means you want to show 
$$3-\frac{1}{a_n}>a_n$$

anonymous · Answer

*write down

anonymous · Answer

@Loser66 for some reason I can't reply to your message. Here it is: Ahh okay that makes sense.  I think we should do it like in Cal2

@satellite73 
Yes...

anonymous · Answer

Increasing means that the following term is bigger than the previous one, for all terms.
Since a1=1 and a2 will be = 3-1=2, that means that a1<a2.

But that doesn't prove the "for all" part, right?  Or is that enough?

anonymous · Answer

no that doesn't prove it for all

anonymous · Answer

I don't know how... :|

anonymous · Answer

maybe induction?

anonymous · Answer

assume that $a_n<3$ and prove that $a_{n+1}<3$ as well

anonymous · Answer

I will read up on that tomorrow morning, can't remember exactly how that works

anonymous · Answer

show it is the case for $n=2$ which you just did
the assume it is true for $n=k$ and then show it is true for $n=k+1$
in other words if you assume $a_k<3$ prove that $a_{k+1}<3$ this should not be hard

anonymous · Answer

since 
$$a_{k+1}=3-\frac{1}{a_k}$$ and we assume that $a_k<3$ making $\frac{1}{a_k}>\frac{1}{3}$ this tells us that $3-\frac{1}{a_k}<3$

anonymous · Answer

Thanks!

Sorry, I need to get to bed now, otherwise tomorrow will be hell.  :)

Cya