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.