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 $I = ]-\infty; -2[~\cup~ ]-2; +\infty[$.
 A: We fill in some of the detail you are having trouble with.  Note that 
$$U_{n+1}=\frac{2U_n+3}{U_n+2}=2-\frac{1}{U_n+2}.\tag{$1$}$$
It is clear that if we start with $U_0=1$, then all the $U_n$ are positive. Then from $(1)$ we can see that $U_n\lt 2$ for all $n$. 
Also, the $U_i$ are increasing. This can be proved by induction, Suppose that $U_{k+1}\gt U_k$. Then 
$$U_{k+2}=2-\frac{1}{U_{k+1}+2}\gt 2-\frac{1}{U_k+2}=U_{k+1}.$$
Thus our sequence is increasing and bounded above, so has a limit. Now we can find the limit by solving a quadratic.
A: $$U_{n+1}-\sqrt3=R_n\cdot(U_n-\sqrt3),\quad R_n=\frac{2-\sqrt3}{U_n+2},\quad 0\lt R_n\leqslant\frac{2-\sqrt3}2\lt1$$
A: This will converge to the fixed point of the map $f$
$$x\mapsto {2x+3\over x + 2}.$$
Set 
$$ x = {2x+3\over x + 2}.$$
and you will get $x = \pm\sqrt{3}$.
The start point $x = 1$ is attracted to $\sqrt{3}$.  
This is the fixed point theorem.  The derivative of our mapping is
$$f'(x) =  {1\over (x + 2)^2}$$
The derivative of this is less than 1 in absolute value for all $x > 0$, so the fixed point will be $\sqrt{3}$.  
This is Picard iteration.
A: To prove existence of the limit put $$f(x) = \frac{2x+3}{x+2}$$ then we want to find the limit of $f^{(n)}(1)$ the $n$th iterate of $f$.
If we show that $f$ contracts some interval containing $1$ we have proved existence of the limit by Banach Fixed Point Theorem, i.e. we need to show $|f(a)-f(b)|<\eta^{-1}|a-b|$ for $a,b \in [1,2]$ and some $\eta^{-1} < 1$. Since $$|f(a)-f(b)| = \frac{|a - b|}{|ab + 2a + 2b + 4|}$$ we just need to show the denominator $> 4 = \eta$, say, which is obviously true.
To calculate what the limit is is equivalent to solving $f(x) = x$ or $\frac{x^2 - 3}{x + 2} = 0$ so it's just $\sqrt{3}$.
Alternatively just take the eigenvalues of the matrix [2,3;1,2] to get the fixed points.
A: We have $U_{n+1}=2-\cfrac{1}{U_n+2}$ and we have $U_1>0$, from which it follows that $\frac32\leq U_n < 2$ for $n>1$, and we can easily refine the estimate.
EDIT
So to proceed differently from André Nicholas, we plug the bounds $\frac 32$ and 2 into the recurrence and obtain a refined estimate for $n>2$ of$$\frac{12}{7}\leq U_n<\frac{7}{4}$$ and then for $n>3$ of$$\frac{45}{26}\leq U_n<\frac{26}{15}$$
This is not the quickest method, but it is exploitable and is intimately connected with the convergents/continued fraction for $\sqrt 3$ since the estimates have the form:
$$\frac{3a}{b}\leq U_n<\frac{b}{a}\text{ with }3a^2=b^2-1$$
André's method is better for the question as asked. I thought this observation was too interesting to leave out.
A: 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 $U=\sqrt{3}$.
