Proof by induction

Assume $1\leq a_n\leq2$ for some $n \in \mathbb{N}$.

Then $a_{n+1} = \frac{2+2*2}{2+1} = 2$ for maximum (need larger numerator and smaller denominator) and $a_{n+1} = \frac{2+2*1}{2+2} = 1$ for minimum (need smaller numerator and larger denominator).

So it is true for n+1 since $1 \leq a_{n+1} \leq 2$.

Done.

You'd also have to explain/prove that it's an increasing sequence.

I would rewrite the fraction

$\displaystyle a_{k+1}=\frac{2+2a_k}{2+a_k} = \frac{4+2a_k-2}{2+a_k} = 2 - \frac{2}{2+a_k}$

Then it's clearer why $1\leq a_{k+1} \leq 2$.

You'd also have to explain/prove that it's an increasing sequence.

Why is that?

I think he's trying to say if i prove it is increasing then I can sub in the end points to show a_{n+1} is bounded between 1 and 2.

Otherwise 1.5 could be used for example etc.

I thought you had justified that with your argument.

How so? By maximising and minimising the denominator (or vice versa)?

It's not really a justification since both the numerator and denominators are functions of $a_n$; so you need to optimise them jointly. (i.e: why is an = 1 the minimum and why is an = 2 the maximum? The answer is because it's an increasing sequence).

For example, $f(x) = \frac{1-x^2}{x+2}$ between $-1 \leq x \leq 1$, it's not true that the minimum occurs at x=-1 and the maximum occurs at x=1 or at x=0 by "maximising the numerator and minimsing the numerator" those arguments don't work without justifying that it's an increasing sequence (which this one isn't)).

Ok, I wasn't claiming that the argument could be generalised, only that it was justifiable in this case.

Only justifiable if it was an increasing sequence, which is precisely what notnek said, that he hadn't justified that it was an increasing sequence.

I don't get what this has to do with whether the sequence is increasing or not.