# Really stumped with a limit proof question

1. Mar 9, 2012

### John765

1. The problem statement, all variables and given/known data

I have the following question that I'm struggling with quite a bit:

Our instructor gave us the hint that we could break it up by proving

a) Proving that 0 <= an < 1 which I managed to do by induction

b) Proving that 0 <= an =< k^(n-1) for n > 0

and then using the squeeze theorem.

2. Relevant equations

None that I know of?

3. The attempt at a solution

I'm having a lot of trouble with part b, I had the thought that I could try to again do it by induction but I'm hitting a bit of a brick wall. I'll show my working so far.

I thought that if I could prove that a(n+1) =< an then it would be trivial to prove by induction that an =< k^(n-1) because I could show a1 =< k^(1-1) and then claim an =< k^(n-1) and then it'd follow that a(n+1) =< k^(n-1) because a(n+1) =< an. Is this a legitimate way of proving it?

So I was thinking I needed to prove that an >= a(n+1) and I tried doing it like this:

so k = a(n+1)/(an(1 - an))

then for an =\= 0:

0 <= a(n+1)/(an(1 - an)) < 1
0 <= a(n+1)/an < (1 - an)
0 <= a(n+1)/an < 1
0 <= a(n+1) < an

So as long as an isn't 0 then it's always larger than a(n+1). This is great and helps me with the k^(n-1) bit.

but what about when an is 0? How on earth do I factor that in? I've been scratching my head for ages but haven't really come up with a way to go about doing it.

My best thought so far is to prove that if an is ever 0 then it has always been 0 and it always will be which means that it will always be equal or less than k^(n - 1). I know an will only be 0 if a0 is equal to 0 or 1 or if k is 0 and I think I can show fairly easily if any of those conditions are true then an will stay 0 forever but how do I go back the other way and prove that if an is 0 it means it always has been 0? (and therefore always less than k^(n-1)?) Really pulling out my hair here.

Or am I going about it completely wrong and there's an easy way of doing this that I'm completely missing?

I get the feeling I might be a very very small nudge away from getting there so if anyone has any thoughts they'd really be appreciated!

2. Mar 9, 2012

### Dick

If you look at the function f(x)=x*(1-x) then what's the maximum of that function for x=[0,1]?

3. Mar 9, 2012

### John765

It has a maximum between 0 and 1 so at x = 0.5 so 0.25. Sorry, wish I was smart enough to see where you were going with this but I'm not sure what to do with this information.

Last edited: Mar 9, 2012
4. Mar 9, 2012

### Dick

Actually, you probably don't even need that. If you have already figured out that 0 <= an < 1. Then isn't an*(1-an)<an? So $a_{n+1}<=k a_n$.

5. Mar 9, 2012

### John765

I'm trying to prove that an*(1-an)<an so that I can prove $a_{n+1}<=k a_n$ but I'm having trouble with the former proof.

6. Mar 9, 2012

### Dick

(1-an)<=1. So (1-an)*an<=an. Isn't that enough?

7. Mar 10, 2012

### John765

What am I doing wrong here?

0 <= 1 - an <= 1
0 <= k*(1 - an) < 1 (because 0 <= k < 1)
0 <= k*an(1 - an) < an
0 <= a(n + 1) < an

I know a(n + 1) can be equal to an so I shouldn't be getting a strictly less than there should I? But there's definitely no way for k*(1 - an) = 1, because k can't be 1.

8. Mar 10, 2012

### Dick

I think the only way a(n+1) can equal a(n) is if a(n)=0. I don't think you are doing anything wrong.

9. Mar 10, 2012

### John765

I agree but shouldn't my inequality reflect that? For me to say that a(n+1) is less than an is correct isn't it? Or do I say that a(n+1) is less than an unless an and a(n+1) are both 0? Wouldn't I need to prove that?

10. Mar 10, 2012

### Dick

If you have the strict inequality a<b and c>=0, then you only know ac<=bc. You can't conclude ac<bc because c could be 0. Is that what you are asking about?