Heron's method for √2

1.
$$u_{n+1}-\sqrt2=\frac{u_n^2+2-2\sqrt2u_n}{2u_n}=\frac{(u_n-\sqrt2)^2}{2u_n}.$$

2. By part 1 and $u_n>0$, $u_{n+1}\ge\sqrt2$ for all $n$. For $n\ge1$,
$$u_{n+1}-u_n=\frac{2-u_n^2}{2u_n}\le0,$$
so $(u_n)$ is decreasing and bounded below by $\sqrt2$: it converges to some $\ell\ge\sqrt2$ with $\ell=\frac12\left(\ell+\frac2\ell\right)$, i.e. $\ell^2=2$, so $\ell=\sqrt2$.

3. With $\varepsilon_n=u_n-\sqrt2$: $\varepsilon_{n+1}=\frac{\varepsilon_n^2}{2u_n}\le\frac{\varepsilon_n^2}{2\sqrt2}\le\varepsilon_n^2$. This is quadratic convergence.

Numerics.
- $u_1=1.5$
- $u_2=1.41667$
- $u_3=1.414216$
- $u_4=1.41421356237\dots$

The number of correct decimals roughly doubles at each step: this is Newton's method applied to $x^2-2=0$.