<s>[INST] limit of convergent series What is the limit of $U_{n+1} = \dfrac{2U_n + 3}{U_n + 2}$ and $U_0 = 1$? 
I need the detail, and another way than using the solution of $f(x)=x$, as $f(x) = \frac{2x+3}{x+2}$ because I can't show that $f(I) \subseteq I$ as $I = ]-\infty; -2[~\cup~ ]-2; +\infty[$.
 [/INST] First note that
$$
\begin{align}
U_{n+1}
&=\frac{2U_n+3}{U_n+2}\\
&=2-\frac1{U_n+2}\tag{1}
\end{align}
$$
Therefore, if $U_n\ge1$, then $\frac53\le U_{n+1}\lt2$.
$$
\begin{align}
U_{n+1}-U_n
&=\frac1{U_{n-1}+2}-\frac1{U_n+2}\\
&=\frac{U_n-U_{n-1}}{(U_n+2)(U_{n-1}+2)}\tag{2}
\end{align}
$$
Since $(U_n+2)(U_{n-1}+2)\ge11$, $(2)$ says that the sequence is monotonic. Thus, because the sequence is monotonic and bounded above and below, it converges to some limit, $U$.
$(1)$ says that
$$
U=\frac{2U+3}{U+2}\Rightarrow U^2=3\tag{3}
$$
Since $U\ge0$, we get that $U=\sqrt{3}$.
</s>