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!