To show sequence $a_{n+1}= \frac{a_n^2+1}{2 (a_n+1)}$ is convergent Let $a_1=0$ 
 and $$a_{n+1}= \dfrac{a_n^2+1}{2 (a_n+1)}$$ $\forall n> 1.$
 Show that sequence $a_n$ convergent.
I tried to prove $a_n$ is less than 1 by looking at few terms. But i failed to prove my claim.
 A: Clearly $a_n\ge 0\forall n.$ $$a_{n+1}= \dfrac{a_n^2+1}{2 (a_n+1)}$$ $$a_n^2-2a_na_{n+1}+(1-2a_{n+1})=0$$ This is a quadratic equation of $a_n.$ Since $a_n\in\mathbb{R}$ we have $$4a_{n+1}^2-4(1-2a_{n+1})\ge 0$$ $$(a_{n+1}+1)^2\ge 2$$ Hence $$a_{n+1}\ge \sqrt{2}-1.$$ Now consider $$a_{n+1}-a_n=\dfrac{-a_n^2-2a_n+1}{2 (a_n+1)}=-\dfrac{(a_n+1+\sqrt{2})(a_n+1-\sqrt{2})}{2 (a_n+1)}\le 0.$$ Therefore $a_n$ is decreasing and bounded below for $n\ge 2$.
A: since
$$a_{n+1}-(\sqrt{2}-1)=\dfrac{a^2_{n}+1-2(\sqrt{2}-1)(a_{n}+1)}{2a_{n}+2}=\dfrac{[a_{n}-(\sqrt{2}-1)]^2}{2a_{n}+2}\tag{1}$$
simaler
$$a_{n+1}+\sqrt{2}+1=\dfrac{(a_{n}+\sqrt{2}+1)^2}{2a_{n}+2}\tag{2}$$
$\dfrac{(1)}{(2)}$ we have
$$\dfrac{a_{n+1}-(\sqrt{2}-1)}{a_{n+1}+\sqrt{2}+1}=\left(\dfrac{a_{n}-(\sqrt{2}-1)}{a_{n}+\sqrt{2}+1}\right)^2=\cdots=\left(\dfrac{a_{0}-(\sqrt{2}-1)}{a_{0}+\sqrt{2}+1}\right)^{2^n}\to 0$$
beause
$$\dfrac{a_{0}-(\sqrt{2}-1)}{a_{0}+\sqrt{2}+1}<1$$
so
$$\lim_{n\to\infty}a_{n}=\sqrt{2}-1$$
A: 1) Clearly (add a little explanation here) the sequence is non-negative for all indexes
2) For $\;n\ge 2\;$ , the sequence is monotonic decreasing, since:
$$a_{n+1}\le a_n\iff \frac{a_n^2+1}{2(a_n+1)}\le a_n\iff a_n^2+2a_n-1\ge 0\iff$$
$$\iff (a_n+1+\sqrt2)(a_n+1-\sqrt2)\ge 0\iff\begin{cases}a_n\le -1-\sqrt2\\{}\\or\\{}\\a_n\ge-1+\sqrt2\end{cases}$$
The first case above is impossible, and the second one follows from
$$a_{n+1}\ge-1+\sqrt2\iff a_n^2+1\ge (2\sqrt2-2)a_n+2\sqrt2-2\iff$$
$$\iff a_n^2-2(\sqrt2-1)a_n-(2\sqrt2-3)\ge 0\iff (a_n+1-\sqrt2)^2\ge 0$$
and we're done since the last inequality is trivially true.
Now you can easily deduce the sequence converges, and its limit $\;x\;$ fulfills
$$x=\frac{x^2+1}{2(x+1)}\implies x=-1+\sqrt2$$
