<s>[INST] 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.
 [/INST] 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
$$\lim_{n\to\infty}a_{n}=\sqrt{2}-1$$
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