Cauchy sequence problem Let $(a_n)_{n\in\mathbb N}$ be defined by the recursive formula $a_1 = 1$, $a_{n+1}=\frac{2 + a_n}{1 + a_n}$ for all $n\geqslant 1$. Show that $(a_n)_{n\in\mathbb N}$ is a Cauchy sequence. 
 A: HINT:$$(*) a_{n+1}=1+\dfrac{1}{1+a_n}>1$$
$$|a_{m+1}-a_{n+1}|=|\dfrac{1}{1+a_m}-\dfrac{1}{1+a_n}|=\\
|\dfrac{a_n-a_m}{(1+a_n)(1+a_m)}|$$ now consider (*)
$$|\dfrac{a_n-a_m}{(1+a_n)(1+a_m)}|<|\dfrac{a_n-a_m}{2\times 2}|\\
|a_{m+1}-a_{n+1}|=|\dfrac{a_n-a_m}{(1+a_n)(1+a_m)}|<\dfrac{1}{4}|a_n-a_m|\\ \to \\|a_{m+1}-a_{n+1}|<\dfrac{1}{4}|a_n-a_m|<\dfrac{1}{4^2}|a_{n-1}-a_{m-1}|\\<\dfrac{1}{4^3}|a_{n-2}-a_{m-2}|<...<\epsilon$$
A: $$a_{n+1}-\sqrt2=\frac{(\sqrt2-1)(\sqrt2-a_n)}{a_n+1}.$$
Thus, $$\left|a_{n+1}-\sqrt2\right|=\frac{\sqrt2-1}{1+a_n}\cdot|a_n-\sqrt2|\leq(\sqrt2-1)|a_n-\sqrt2|,$$
which says that your sequence  converges to $\sqrt2$ and from here it's the Cauchy sequence.
A: $f(x)=\frac{2+x}{1+x}$ is a positive and decreasing function on $\mathbb{R}^+$.  The only positive solution of $f(x)=x$ is given by $x=\sqrt{2}$ and
$$\forall x\in\left[1,\frac{3}{2}\right],\qquad \left|f'(x)\right| = \frac{1}{(x+1)^2}\leq \frac{1}{4}. $$
By the Banach fixed point theorem it follows that $\{a_n\}_{n\geq 1}$ is convergent to $\sqrt{2}$, hence it is a Cauchy sequence.
