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.

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 $a_{\infty}$.
