Limit of a recursively defined sequence

Question

$ a_{n+1} = \frac{3+2a_n}{3+a_n} $ and $ a_0 = 1 $

This sequence is obviously increasing, so if we could prove it is bounded, we'd also prove it converges and we could easily find the limit by $ L = \frac{3+2L}{3+L}$, but I'm somewhat clueless about how to prove that it indeed is bounded.

Answer 1

It is pretty obviously bounded... just look at the numerator and denominator of the fraction. I don't want to give away the answer just yet, because I am sure if you look closely you can find an easy upper bound.

Answer 2

If $1\lt a_n\lt2$, then

$$1={3+2\cdot1\over3+2}\lt{3+2a_n\over3+a_n}\lt{3+2\cdot2\over3+1}={7\over4}\lt2$$

Alternatively (and more directly),

$$1={3+a\over3+a}\lt{3+2a\over3+a}\lt{6+2a\over3+a}=2$$

for any $a\gt0$. But what's problematic is the OP's claim that the sequence is "obviously" increasing. Looked at more closely,

$$a\le{3+2a\over3+a}\iff a^2+a-3\le0$$

which is only true in the interval $\left[{-1-\sqrt{13}\over2},{-1+\sqrt{13}\over2}\right]\approx[-2.303,1.303]$

So one really needs to prove the $a_n$'s stay in this range.