proving convergence for a sequence defined recursively

Hint: Find $c$ such that $$ \frac{a_{n+1}-\sqrt2}{a_{n+1}+\sqrt2}=c\,\frac{a_{n}-\sqrt2}{a_{n}+\sqrt2}. $$

The function $\displaystyle f(x) = 1 + \frac{1}{1+x}$ has derivative $\displaystyle f'(x) = - \frac{1}{(1+x)^2}.$ On $[1,\infty)$ we have $-1/4 \leq f'(x) <0$ so the sequence converges by the Contraction Mapping Theorem.

This answer shows that if $b$ and $d$ have the same sign, then $$ \frac{a+c}{b+d}\text{ is between }\frac{a}{b}\text{ and }\frac{c}{d} $$ Therefore, $$ a_{n+1}=\frac{a_n+2}{a_n+1}\tag{1} $$ is between $1$ and $2$.

Furthermore, $$ \begin{align} a_{n+1}^2-2 &=\left(\frac{a_n+2}{a_n+1}\right)^2-2\\ &=\frac{2-a_n^2}{(a_n+1)^2}\tag{2} \end{align} $$ Therefore, induction and $(2)$ show that $$ 0\le(-1)^n(2-a_n^2)\le\frac1{4^n}\tag{3} $$ Thus, $(3)$ shows that $$ \lim_{n\to\infty}a_n=\sqrt{2}\tag{4} $$