How can I show the convergence and find limit. 
Let $a_1=0$ and $$a_{n+1}=\frac{a_n^2+3}{2(a_n+1)}~,~n \geq 1$$ . Show that $\{\ a_n \}$ converges  and find its limit.

Trial: Here $$\begin{align} a_1 &=0\\ \text{and}~ a_2 &=\frac{3}{2} \\ \text{and}~ a_3 &=\frac{21}{20} \end{align}$$ . Here I find no relationship between $a_n ~\text{and}~ a_{n+1}$. Please help.
 A: I first saw the following slick method in Bartle and Sherbert's Introduction to Real Analysis (the slickness due to them, not me).
Prove that $(a_n)$ is bounded below and eventually decreasing as follows:
1) Using the relation $a_n^2-(2a_{n+1})a_{n}+(3-2a_{n+1})=0$ and the Quadratic formula, show that $a_{n+1}\ge1$ for $n>1$.
2) Use the result of 1) to show that the quantity $a_n-a_{n+1}$ is non-negative for $n>1$.
Of course, then, it follows that $(a_n)$ converges, and you can find the value of the limit using your recurrence relation.
The above is meant to be a (big) hint).  The details are given below.




From the definition of $a_{n+1}$, you have
$$
a_n^2-(2a_{n+1})a_{n}+(3-2a_{n+1})=0.
$$
So, $a_{n}$ is a solution of the quadratic equation $x^2-(2a_{n+1})x +(3-2a_{n+1})=0$.
We must then have
$$
(2a_{n+1})^2-4\cdot 1\cdot(3-2a_{n+1})\ge0.
$$
This implies that $$a_{n+1}^2 + 2a_{n+1}\ge3.$$
And then, as $a_{n+1}>0$, it follows that  $a_{n+1}\ge1$ for $n>1$.
Now, consider 
$$
a_n-a_{n+1} = a_n-{a_n^2+3\over 2(a_n+1)} ={(a_n+3)(a_n-1)\over 2(a_n+1)}.
$$
Since $a_n\ge1$ for $n>1$, we have $a_n-a_{n+1}\ge 0$ for $n>1$.
We have shown that $(a_n)$ is an eventually decreasing sequence that is bounded below.  It thus converges to some   $L$. Using your recurrence relation, you can find the value of $L$.
A: Okay what do we need for conergence? If we can show that the sequence is bounded and monotone we know that it converges.


*

*bounded: Obviously we have $0\leq a_n$. I will show that $a_n<10$ (which is rather arbitary) by induction. We have $a_0<10$. Now we make the step $n\rightarrow n+1$.
\begin{align} a_{n+1} &= \frac{a_n^2+3}{2(a_n+1)} <\frac{a_n^2+2a_n+3}{2(a_n+1)}
\\ &=\frac{(a_n+1)^2+2}{2(a_n+1)} = \frac{(a_n+1)^2}{2(a_n+1)}+\frac{2}{2(a_n+1)}
\\ &= \frac{a_n+1}{2} + \frac{2}{2(a_n+1)} \leq^* \frac{10+1}{2}+\frac{2}{2(0+1)} 
\\ &= 5.5 +1 <10\end{align}
Where we use $0\leq a_n<10$ at $^ *$.

*monotone: here we show, that $\forall n > 0$ we have $\displaystyle \frac{a_{n+1}}{a_n}<1$. I think this is something you should proof.


For the last step: We  proofed that our sequence converges.
So there exists an $a=\lim_{n\rightarrow \infty} a_n$ Knowing this, you can solve
$a_{\infty} = \frac{a_{\infty}^2+3}{2(a_{\infty}+1)}$ for $a_{\infty}$.
A: $a_1=0 \Rightarrow a_2= \frac{0^{2}+3}{2(0+1)}=\frac{3}{2}$. Note that $\frac{3}{2}\geqslant\frac{3}{2}>1$.
Next, $a_n>1\Rightarrow \exists k>0$ such that $a_n = 1+k$.
Also, if $k\leqslant \frac{1}{2}$, then
$a_{n+1}=\frac{(1+k)^{2}+3}{2((1+k)+1)}=\frac{4+2k+k^{2}}{4+2k}=1+\frac{k^{2}}{4+2k}(*)<1+\frac{k^{2}}{4}+\frac{k^{2}}{2k}=1+\frac{k}{2}(1+\frac{k}{2})$
$\leqslant 1+\frac{k}{2}(1+\frac{1}{4})=1+\frac{k}{2}(\frac{5}{4})=1+ \frac{5}{8}k<1+k=a_n.$
I noticed the inequality at $(*)$ works only after I deleted this answer, because at first it was a genuine schoolboy error, but it actually is true and therefore I can luckily salvage this answer. Why is $(*)$ true? Because $a>0,b>0,c>0\Rightarrow a<2a=\frac{ab}{b}+\frac{ac}{c}<\frac{a(b+c)}{b}+\frac{a(b+c)}{c} \ $. Now divide both sides by $(b+c)$: $\quad \frac{a}{b+c}<\frac{a}{b}+\frac{a}{c}. \ $ In fact, we could make the restriction of k above even stronger: $1+\frac{k^{2}}{4+2k}<1+ \frac{5}{16}k \ $ because $\quad \frac{a}{b+c}<\frac{1}{2}(\frac{a}{b}+\frac{a}{c}), \ $ however I have already used $\frac{5}{8}$ throughout the rest of the proof and it suffices, so let's stick with that.
From this, we see that:
$$\bbox[yellow]{(1)\qquad 1<a_{n+1}<1+\frac{5}{8}k<1+k=a_n\leqslant\frac{3}{2}\qquad for\ all \ n\geqslant 2.}\ $$
From $(1)$ we deduce: {$a_n$} is monotonically decreasing and converges to some limit $L$ where $\frac{3}{2}\geqslant L\geqslant1$, and also $$\bbox[yellow]{(2)\quad\qquad\qquad\qquad\qquad a_{n+1}-1<\frac{5}{8}(a_n-1) \qquad for\ all \ n\geqslant 2,}\ $$
because $k=a_n-1.$
$(2)$ says that the distance between $a_{n+1}$ and 1 is $<\frac{5}{8}$ of the distance between $a_n$ and 1.
Thus $$\bbox[yellow]{(3)\quad\qquad\qquad\qquad\qquad 1<a_n\leqslant 1+\frac{1}{2}(\frac{5}{8})^{n-2} \qquad for\ all \ n\geqslant 2.}\ $$
At this stage, I would invoke the Sandwich Theorem. Our sequence {$a_n$} is "squeezed" or "sandwiched" by the two sequences {$b_n$} = {$1$} and {$c_n$} = {$1+\frac{1}{2}(\frac{5}{8})^{n-2}$}, both of which converge to $1$. Therefore, by the Sandwich Theorem for sequences, $\ L=1$.
Instead of invoking the Sandwich Theorem, you could also go the "formal route".
Given ${\varepsilon}>0,$ no matter how small, we need to find $N$, in terms of ${\varepsilon}$, such that $a_n-1<{\varepsilon} \ $ for all $n \geqslant N$
$a_n-1<{\varepsilon} \iff a_n \leqslant 1+\frac{1}{2}(\frac{5}{8})^{n-2}<1+{\varepsilon} \iff (\frac{5}{8})^{n-2}<2{\varepsilon} \iff (n-2)log(\frac{5}{8}) <log(2{\varepsilon}) \iff n-2> \frac{log(2{\varepsilon})}{log(5/8)} \iff$
$$\bbox[yellow]{n>\frac{log(2{\varepsilon})}{log(5/8)}+2}\ , $$
and so we have found $N({\varepsilon})$, the least integer greater than $\frac{log(2{\varepsilon})}{log(5/8)}+2$, which guarantees $a_n-1<{\varepsilon}$ for all $n\geqslant N.$
