How can I show the sequence $a_{n+1}=\frac{4a_{n}+2}{a_{n}+3}$ is bounded for every $a_{1}\in \mathbb{R}$? Investigating this sequence led me to the following conclusions:


*

*if $a_{1}\geq 2$ then $a_{n}\geq 2$ and $a_{n}$ is decreasing, thus it converges and is bounded.

*if $-1\lt a_{1} \lt 2$ then $a_{n}$ is increasing, thus it converges and is bounded.

*if $a_{1}=-1$ or $a_{1}=2$ the sequence is constant and is bounded.

*if $a_{1}\lt -1$ then for some $k,~a_{k}\gt 0$ and then we're in one of the previous options.


But this seems overly complicated and very long to prove.
I thought of trying the following argument: define $f(x)=\frac{4x+2}{x+3}$. We see that $\displaystyle \lim_{x\to \infty}f(x)=4$ and $f(x)$ is continuous in the interval $[0, \infty)$ therefore  it's bounded there. Then we show that if $a_{1}<0$ then $\exists N,~\forall n\gt N,~a_{n}>0$ which gives us $a_{n+1}=f(a_{n})$ is bounded.
 A: Write $a_n={2+b_n \over 1-b_n}$. Then $b_{n+1}={3\over5} b_n$. This implies $\lim_{n\to\infty} b_n=0$ and $\lim_{n\to\infty} a_n=2$, unless $b_n=1$ for some $n\ge0$. This is the case if $b_0=(5/3)^k$ for some $k\ge0$. In addition there is the fixed point $a_0=a_1=\ldots=-1$ which is not covered by this "Ansatz".
A: You can formalize what you're doing by showing that $-1 \leq a_{n+1} < 4$ whenever $a_n \geq -1$. So once you get an $n$ such that $a_n \geq -1$ you're done.
If some $a_n < -3$, your recursion gives $a_{n+1} > -1$ and you're in the above situation. The only remaining option is that $-3 < a_n < -1$ for all $n$... but then the sequence is still bounded. So you're done.
A: What about $a_1 = -3$? The function don't even exists there.
Anyway, your argument is valid: if $a_n = f(n)$ and $\lim_{x\to \infty} f(x) = L$ then $\lim_{n\to \infty} a_n = L$.
You never care about the first terms of the sequence, because when you've convergence to the infinity, you can just pick a maximum value and set it as limit. For example, if your sequence converges to 4, you can set a bound as $\max (4, a_1, a_2, ..., a_k)$ for any k you like.
A: Your last argument is sound.  If you know that the terms are bounded over $[0,\infty)$, (or any interval, for that matter) that iteration doesn't take it out of that interval, and that all starting values eventually enter that interval, the terms are bounded.  As AkiRoss says, prepending terms cannot change the bounded nature, it only changes the bound.
