Can Induction and the Squeeze Theorem Solve This Limit Problem?

[John765]

Homework Statement

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 <= a_n < 1 which I managed to do by induction

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

and then using the squeeze theorem.

Homework Equations

None that I know of?

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

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

so k = a_(n+1)/(a_n(1 - a_n))

then for a_n =\= 0:

0 <= a_(n+1)/(a_n(1 - a_n)) < 1
0 <= a_(n+1)/a_n < (1 - a_n)
0 <= a_(n+1)/a_n < 1
0 <= a_(n+1) < a_n

So as long as a_n 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 a_n 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 a_n 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 a_n 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 a_n will stay 0 forever but how do I go back the other way and prove that if a_n 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!

 

[Dick]

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

 

[John765]

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

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.

 

[Dick]

It has a maximum between 0 and 1 so at x = 0.5 so 0.375. 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.

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}&lt;=k a_n.

 

[John765]

Then isn't an*(1-an)<an? So a_{n+1}&lt;=k a_n.

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

 

[Dick]

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

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

 

[John765]

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

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.

 

[Dick]

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.

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.

 

[John765]

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.

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?

 

[Dick]

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?

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?