(adsbygoogle = window.adsbygoogle || []).push({}); 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 <= 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.

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)}=< a_{n}then it would be trivial to prove by induction that a_{n}=< k^(n-1) because I could show a_{1}=< 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 a_{0}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!

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# Homework Help: Really stumped with a limit proof question

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