$a_1 =2$ and $a_{n+1}= \frac{2a_n +3}{a_n +2}$ the recursive sequence , converges?, and if yes, show it to which converges I do not know what to do, I try showing the first elements to see if they were behaving in some way, but no

Suppose the sequence converges to $a$, then $a(a+2)=2a+3$, that is $a=\sqrt3$ (if we have $a_n\gt0$, then the recursion guarantees that $a_{n+1}\gt0$).
To show convergence, note that
$$
\begin{align}
a_{n+1}-\sqrt3
&=\frac{2a_n+3}{a_n+2}-\sqrt3\\
&=\frac{(2-\sqrt3)(a_n-\sqrt3)}{a_n+2}
\end{align}
$$
Since $a_n\ge0$, we have that
$$
\left|a_{n+1}-\sqrt3\right|\le\tfrac{2-\sqrt3}{2}\left|a_n-\sqrt3\right|
$$
Therefore,
$$
\left|a_n-\sqrt3\right|\le\left(\tfrac{2-\sqrt3}2\right)^{n-1}\left|a_1-\sqrt3\right|
$$
Since $\left|\frac{2-\sqrt3}{2}\right|\lt1$, the sequence converges to $\sqrt3$ for any positive $a_1$.
