Approximating √2 by 1 + 1/(1 + x)

Consistency. $1+\frac1{1+p/q}=1+\frac q{p+q}=\frac{p+2q}{p+q}$, so $x_n=\frac{p_n}{q_n}$ indeed.

1. Since $x_{n+1}=\frac{2+x_n}{1+x_n}$,
$$x_{n+1}-\sqrt2=\frac{2+x_n-\sqrt2-\sqrt2x_n}{1+x_n}=\frac{(1-\sqrt2)(x_n-\sqrt2)}{1+x_n}.$$
All $x_n\ge1$, so
$$|x_{n+1}-\sqrt2|\le\frac{\sqrt2-1}{2}|x_n-\sqrt2|<\frac{|x_n-\sqrt2|}{4}.$$
Hence $|x_n-\sqrt2|\le4^{-n}|x_0-\sqrt2|\to0$: geometric convergence.

2.
$$p_{n+1}^2-2q_{n+1}^2=(p_n+2q_n)^2-2(p_n+q_n)^2=-p_n^2+2q_n^2=-(p_n^2-2q_n^2).$$
The sign alternates, and $p_0^2-2q_0^2=-1$, so $p_n^2-2q_n^2=(-1)^{n+1}$.

3. $\left|\sqrt2-\frac{p_n}{q_n}\right|=\dfrac{|2q_n^2-p_n^2|}{q_n^2\left(\sqrt2+\frac{p_n}{q_n}\right)}=\dfrac1{q_n^2\left(\sqrt2+x_n\right)}<\dfrac1{2q_n^2}$, since $\sqrt2+x_n>\sqrt2+1>2$.

First terms. $1,\ \frac32,\ \frac75,\ \frac{17}{12},\ \frac{41}{29},\dots$ and $3^2-2\cdot2^2=1$, $7^2-2\cdot5^2=-1$, $17^2-2\cdot12^2=1$.

Lesson. An iteration $x\mapsto\frac{ax+b}{cx+d}$ with a fixed point $\alpha$ contracts: compute $x_{n+1}-\alpha$ as a multiple of $x_n-\alpha$.