$a_1 =2$ 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

Another less known way of solving this particular form of recurrence relations.
The key is transform the non-linear recurrence to a linear one.
Notice $$\begin{bmatrix}a_{n+1}\\1\end{bmatrix}
\propto \begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}
\begin{bmatrix}a_{n}\\1\end{bmatrix}, \forall n > 1
\quad\text{ and }\quad
\begin{bmatrix}a_{1}\\1\end{bmatrix} = \begin{bmatrix}2\\1\end{bmatrix} =
\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}
\begin{bmatrix}1\\0\end{bmatrix}$$
We have
$$\begin{bmatrix}a_{n}\\1\end{bmatrix}
\propto \begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}^n
\begin{bmatrix}1\\0\end{bmatrix}, \forall n > 1$$
The matrix $\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}$ has eigenvalues $\lambda_{\pm}$ and corresponding eigenvectors $u_{\pm}$ given by
$$\lambda_{\pm} = 2\pm \sqrt{3}\quad\leftrightarrow\quad u_{\pm} = \begin{bmatrix}\pm\sqrt{3}\\1 \end{bmatrix}$$
Since $\begin{bmatrix}1\\0\end{bmatrix} \propto u_{+} - u_{-}$, we have
$\begin{bmatrix}a_{n}\\1\end{bmatrix} \propto \lambda_{+}^n u_{+} - \lambda_{-}^n u_{+}$. Together with $\lambda_{+}\lambda_{-} = 1$ and $|\lambda_{-}| < 1$, we find
$$a_n = \sqrt{3}\frac{\lambda_{+}^n - \lambda_{-}^n}{\lambda_{+}^n - \lambda_{-}^n} =
\sqrt{3}\frac{1 + \lambda_{-}^{2n}}{1 - \lambda_{-}^{2n}}
= \sqrt{3}\frac{1 + (2-\sqrt{3})^{2n}}{1 - (2-\sqrt{3})^{2n}}
\to \sqrt{3}\quad\text{ as } n \to \infty$$
