$a_1 =2$ 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
 A: Let $f(x)=(2\,x+3)/(x+2)$. There is a unique point $x>0$ such that $f(x)=x$, namely $x=\sqrt3$. Since $2>\sqrt3$, I will focus on the interval $[\sqrt3,\infty)$. On that interval $f(x)\ge\sqrt3$ and $f(x)\le x$, with equality only for $x=\sqrt3$. Use this to show that $a_n$ is decreasing and bounded below by $\sqrt3$. This implies that $a_n$ is convergent. Take limits in the equality defining $a_{n+1}$ and find the limit.
A: 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 $a_1$.
A: First we observe that 
$$a_{n+1}-a_n=\frac{3-a_n^2}{2+a_n}$$
For $0<a_0<\sqrt{3}$, we can show by induction that the sequence is positive and increasing while for $a_0>\sqrt{3}$ we can show by induction that the sequence is positive and decreasing.
Next, we write 
$$a_{n+1}=2-\frac{1}{a_n+2}$$
Obviously, for $a_0>0$ we have that $0<a_{n}<2$ for all $n$.  Since the sequence is bounded above and below, and is monotonically increasing for $0<a_0<\sqrt{3}$ and monotonically decreasing for $a_0>\sqrt{3}$,  we have established convergence.  
Denote the limit of the sequence $L$ so that $\lim_{n\to \infty}a_n=L>0$.  Then, we have $$L=2-\frac1{L+2}\implies L=\sqrt{3}$$and we are done!
A: Let $\varphi(x) = \frac{2x+3}{x+2}$. Note that $\varphi(x)$ maps $[1, 2]$ into itself (checking endpoints and monotonicity):
$$\begin{aligned}
\varphi(1) &= \frac{5}{3} \in [1,2]\\
\varphi(2) &= \frac{7}{4} \in [1,2]\\
\varphi'(x) &= \frac{1}{(2+x)^2} > 0, \quad x > 1
\end{aligned}
$$
Also, due to
$$
|\varphi'(x)| \leq q = \frac{1}{9} < 1, \quad x \in [1, 2]
$$
$\varphi(x)$ is a сontraction mapping, thus there exists a fixed point 
$$
x = \varphi(x), \quad x = \sqrt{3} \in [1, 2].
$$
and $x_n \to x$ and
$$
|x_n - x| \leq C q^n
$$
A: 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
$$a_n = \sqrt{3}\frac{\lambda_{+}^n - \lambda_{-}^n}{\lambda_{+}^n - \lambda_{-}^n} =
\sqrt{3}\frac{1 + \lambda_{-}^{2n}}{1 - \lambda_{-}^{2n}}
= \sqrt{3}\frac{1 + (2-\sqrt{3})^{2n}}{1 - (2-\sqrt{3})^{2n}}
\to \sqrt{3}\quad\text{ as } n \to \infty$$
